CP-2 Complex numbers — revision question pack

11 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section CP-2. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

CP-2.1 · Solve any quadratic equation with real coefficients. Solve cubic or quartic equations with real coefficients.

Explanation

A quadratic with real coefficients may have two real roots, one repeated real root or a complex-conjugate pair, determined by the sign of $b^2-4ac$ . Use factorisation, completing the square or $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ , with $\sqrt{-q}=i\sqrt q$ for $q>0$ .
For a cubic or quartic, sufficient information will identify a root or factor.
Apply the factor theorem or polynomial division, then solve every remaining quadratic factor.
A real polynomial cannot have an isolated non-real root: its conjugate is also a root.
A complete solution must list all roots, including repeated and non-real roots, rather than stopping after the supplied factor is used.

Worked example

Given that $z=1$ is a root of $z^3-5z^2+17z-13=0$ , solve the equation completely.

1.The factor theorem gives the factor $z-1$ .
2.Polynomial division gives $z^3-5z^2+17z-13=(z-1)(z^2-4z+13)$ .
3. $z^2-4z+13=(z-2)^2+9$ , so $(z-2)^2=-9$ .
4.Hence $z=1$ or $z=2\pm3i$ .

Answer: $z=1$ , $z=2+3i$ or $z=2-3i$ .

Common mistakes

Don't fall into the trap of stopping after finding the supplied real root and leaving the remaining quadratic unsolved.
Don't fall into the trap of writing $\sqrt{-36}=-6$ instead of $\sqrt{-36}=6i$ .
Don't fall into the trap of losing a repeated root when reporting the complete solution of a quartic.

Exam tip

For a 'solve completely' question, show the factor division and state every real and non-real root.

Tier 1 · Easy

1.
Solve $x^2-4x+13=0$ , giving the roots in the form $a+ib$ .
(3)
(Total for Question 1 is 3 marks)
2.
Solve $2x^2+3x+5=0$ in the form $u+iv$ , where $u$ and $v$ are exact real numbers.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Given that $x=2$ is a root of $x^3-2x^2+5x-10=0$ , solve the equation completely.
(4)
(Total for Question 1 is 4 marks)
2.
For real $t$ , the equation $z^2-2tz+t^2+4=0$ has a root of modulus $5$ . Determine the two possible values of $t$ and the corresponding roots.
(5)
(Total for Question 2 is 5 marks)
3.
The polynomial $p(z)=z^3+az^2-z-33$ , where $a\in\mathbb R$ , has the root $-2+i\sqrt7$ . Find the value of $a$ and solve $p(z)=0$ completely.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Given that $x^2+3x+6$ is a factor of $x^4+3x^2+36$ , solve $x^4+3x^2+36=0$ completely.
(5)
(Total for Question 1 is 5 marks)
2.
For real constants $c$ and $d$ , let $q(z)=z^4+cz^3+dz^2+4cz+68$ . Given that $q(1+4i)=0$ , find the value of $c$ and the value of $d$ , and solve $q(z)=0$ completely.
(7)
(Total for Question 2 is 7 marks)
3.
The polynomial $f(z)=z^4+uz^3+vz^2+166z+408$ , where $u,v\in\mathbb R$ , has the root $3+5i$ . Solve $f(z)=0$ completely, and find the value of $u$ and the value of $v$ .
(8)
(Total for Question 3 is 8 marks)
4.
Solve $z^4+6z^2+25=0$ completely. Hence solve $25w^4+6w^2+1=0$ completely.
(8)
(Total for Question 4 is 8 marks)
5.
Express $z^4-16z^3+146z^2-656z+1681$ as the square of a monic quadratic. Hence solve $z^4-16z^3+146z^2-656z+1681=0$ , stating the multiplicity of each root, and find the sum of the reciprocals of the four roots counted with multiplicity.
(8)
(Total for Question 5 is 8 marks)

CP-2.2 · Add, subtract, multiply and divide complex numbers in the form x + iy with x and y real. Understand and use the terms 'real part' and 'imaginary part'.

Explanation

For $z=x+iy$ , where $x,y\in\mathbb R$ , the real part is $\operatorname{Re}(z)=x$ and the imaginary part is $\operatorname{Im}(z)=y$ ; the imaginary part is not $iy$ . Its modulus is the non-negative length $|z|=\sqrt{x^2+y^2}$ and an argument is its directed angle from the positive real axis.
Addition and subtraction combine corresponding parts.
Multiplication uses ordinary expansion followed by $i^2=-1$ .
To divide by $c+di$ , multiply numerator and denominator by its conjugate $c-di$ so the denominator becomes $c^2+d^2$ , then collect the result into Cartesian form.
Equality of complex numbers means equality of their real parts and equality of their imaginary parts, producing two real equations.

Worked example

Find real numbers $x$ and $y$ if $(2-i)(x+iy)=7+4i$ .

1.Expand: $(2-i)(x+iy)=(2x+y)+i(2y-x)$ .
2.Equate real parts to obtain $2x+y=7$ .
3.Equate imaginary parts to obtain $2y-x=4$ .
4.Solving simultaneously gives $x=2$ and $y=3$ .

Answer: $x=2$ and $y=3$ .

Common mistakes

Don't fall into the trap of stating $\operatorname{Im}(x+iy)=iy$ instead of the real coefficient $y$ .
Don't fall into the trap of using $i^2=1$ when expanding a product.
Don't fall into the trap of dividing real and imaginary parts separately instead of rationalising with the conjugate.

Exam tip

After any Cartesian calculation, collect the result as $x+iy$ before reading off or equating its parts.

Tier 1 · Easy

1.
Let $z=(3-2i)+(5+7i)$ . Find $z$ , $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ .
(2)
(Total for Question 1 is 2 marks)
2.
Given $(a+bi)^2=5+12i$ , where $a$ and $b$ are positive real numbers, find the values of $a$ and $b$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Express $\dfrac{4+i}{2-i}$ in the form $x+iy$ .
(3)
(Total for Question 1 is 3 marks)
2.
Let $z=x+iy$ , where $x,y\in\mathbb R$ . Given that $\operatorname{Re}((2+i)z)=1$ and $\operatorname{Im}\left(\dfrac{z}{1-i}\right)=2$ , determine $z$ .
(4)
(Total for Question 2 is 4 marks)
3.
Write $z=x+iy$ for real $x$ and $y$ . Solve $(3-2i)z+(2+i)\overline z=1+3i$ , giving $z$ in Cartesian form.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Express $\dfrac{(2+3i)(1-4i)}{3+i}$ in the form $x+iy$ .
(5)
(Total for Question 1 is 5 marks)
2.
Complex numbers $z$ and $w$ satisfy $z+(1-i)w=4+3i$ and $iz+(2+i)w=-4+7i$ . Determine $z$ and $w$ in Cartesian form.
(5)
(Total for Question 2 is 5 marks)
3.
Express $\displaystyle\frac{(1+2i)^3}{(3+2i)(1-i)}$ in Cartesian form.
(5)
(Total for Question 3 is 5 marks)

CP-2.3 · Understand and use the complex conjugate. Know that non-real roots of polynomial equations with real coefficients occur in conjugate pairs.

Explanation

The complex conjugate of $z=x+iy$ is $\overline z=x-iy$ , obtained by reflecting the point for $z$ in the real axis. Important identities are $z+\overline z=2\operatorname{Re}(z)$ and $z\overline z=x^2+y^2=|z|^2$ , which is real.
If a polynomial has real coefficients and $a+ib$ is a non-real root, conjugating the whole equation shows that $a-ib$ is also a root.
The pair produces the real quadratic factor $[x-(a+ib)][x-(a-ib)]=(x-a)^2+b^2$ .
This rule depends on the coefficients being real.
In polynomial questions, use the conjugate root before dividing or comparing coefficients to find the remaining real factor.

Worked example

A monic cubic with real coefficients has roots $3+2i$ , $3-2i$ and $-1$ . Find the polynomial in expanded form.

1.The conjugate pair gives $(x-3-2i)(x-3+2i)=(x-3)^2+4$ .
2.This simplifies to $x^2-6x+13$ .
3.Include the third factor: $(x^2-6x+13)(x+1)$ .
4.Expansion gives $x^3-5x^2+7x+13$ .

Answer: $x^3-5x^2+7x+13$ .

Common mistakes

Don't fall into the trap of replacing $a+ib$ by $-a-ib$ instead of by its conjugate $a-ib$ .
Don't fall into the trap of assuming conjugate pairing when the polynomial coefficients are not all real.
Don't fall into the trap of expanding the pair as $(x-a)^2-b^2$ instead of $(x-a)^2+b^2$ .

Exam tip

When a real-coefficient polynomial has one non-real root, write its conjugate and the resulting real quadratic factor immediately.

Tier 1 · Easy

1.
For $z=5-4i$ , write down $\overline z$ and calculate $z\overline z$ .
(2)
(Total for Question 1 is 2 marks)
2.
A monic quadratic with real coefficients has coefficient of $x$ equal to $6$ . One root is non-real and has imaginary part $5$ . Determine both roots and the quadratic.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The polynomial $p(x)=x^3-x^2-7x+15$ has real coefficients, and $2+i$ is a root. Solve $p(x)=0$ completely.
(4)
(Total for Question 1 is 4 marks)
2.
The polynomial $x^3+ax^2+bx+50$ , where $a,b\in\mathbb R$ , has the root $4+3i$ . Find the values of $a$ and $b$ and the other roots.
(5)
(Total for Question 2 is 5 marks)
3.
The monic quadratic $z^2+mz+n$ , where $m,n\in\mathbb R$ , has the root $5-2i$ . Find the values of $m$ and $n$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
A monic quartic polynomial has real coefficients. Two of its roots are $5+i$ and $-1+3i$ . Determine the polynomial in expanded form.
(6)
(Total for Question 1 is 6 marks)
2.
The roots with positive imaginary part of a monic quartic with real coefficients are $a+i$ and $b+4i$ , where $a,b\in\mathbb R$ . Given that $a+b=-1$ , $a^2+b^2=5$ and $a>0$ , write the quartic as $(x^2+px+q)(x^2+rx+s)$ , where $p$ , $q$ , $r$ and $s$ are integers, and state the four roots.
(6)
(Total for Question 2 is 6 marks)
3.
A monic quartic with real coefficients has roots $a\pm2i$ and $b\pm5i$ , where $a,b\in\mathbb R$ and $a-2b<0$ . The coefficient of $x^3$ is $-6$ and the coefficient of $x^2$ is $18$ . Determine the quartic in expanded form and state its four roots.
(7)
(Total for Question 3 is 7 marks)

CP-2.4 · Use and interpret Argand diagrams.

Explanation

An Argand diagram represents $z=x+iy$ by the point $(x,y)$ : the horizontal axis is the real axis and the vertical axis is the imaginary axis.
Addition of complex numbers is vector addition, so the point representing $z_1+z_2$ is obtained by completing a parallelogram.
The difference $z_2-z_1$ is the directed displacement from the point representing $z_1$ to the point representing $z_2$ , and $|z_2-z_1|$ is the corresponding distance.
Coordinate geometry can prove gradients, perpendicularity, lengths and areas.
A labelled sketch should show exact complex numbers or coordinates, but conclusions such as a right angle or equal length still require algebraic justification.

Complex addition shown as vector addition on an Argand diagram.

Worked example

The point $A$ is represented by the complex number $a=2+i$ , and the point $B$ is represented by $b=5+5i$ . Find $|b-a|$ and the complex number represented by the midpoint of $AB$ .

1. $b-a=(5+5i)-(2+i)=3+4i$ .
2. $|b-a|=\sqrt{3^2+4^2}=5$ .
3.The complex number represented by the midpoint is $\dfrac{a+b}{2}=\dfrac{7+6i}{2}$ .

Answer: $AB=5$ and the midpoint is represented by $\dfrac72+3i$ .

Common mistakes

Don't fall into the trap of plotting $x+iy$ at $(y,x)$ by interchanging the real and imaginary axes.
Don't fall into the trap of using $|z_2+z_1|$ for the distance between two represented points.
Don't fall into the trap of claiming a geometric property from an approximate sketch without an algebraic calculation.

Exam tip

For an Argand-diagram distance, form the difference of the two represented complex numbers first and then take its modulus.

Tier 1 · Easy

1.
The complex number $z=-2+3i$ is represented by $P$ on an Argand diagram. State the coordinates of $P$ and the quadrant in which it lies.
(2)
(Total for Question 1 is 2 marks)
2.
The points $A$ , $B$ and $C$ are represented by the complex numbers $1+2i$ , $4+i$ and $5+4i$ , respectively. They are consecutive vertices of parallelogram $ABCD$ . Find the complex number represented by $D$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The points $A$ , $B$ and $C$ are represented by the complex numbers $1+i$ , $5+i$ and $1+4i$ , respectively. Determine the exact area of triangle $ABC$ .
(3)
(Total for Question 1 is 3 marks)
2.
The points $A$ , $B$ and $C$ are represented by the complex numbers $a$ , $b$ and $c$ respectively, where $a=1+i$ and $b=5+2i$ . The point $G$ is represented by $g=3+3i$ , and $a+b+c=3g$ . Determine the complex number represented by $C$ , and find $|c-a|$ and $\arg(c-a)$ .
(4)
(Total for Question 2 is 4 marks)
3.
The points $P$ , $Q$ and $R$ are represented by $-2+3i$ , $10-5i$ and $7-3i$ , respectively. Given that $R$ lies on the line segment $PQ$ , find the ratio $PR:RQ$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
The point $A$ is represented by the complex number $3+i$ , the point $B$ is represented by $-1+3i$ , and $O$ is the origin. Prove that triangle $OAB$ is right-angled and isosceles, and find its exact area.
(5)
(Total for Question 1 is 5 marks)
2.
The points $A$ , $B$ and $C$ are represented by the complex numbers $1+2i$ , $5$ and $t+4i$ , respectively, where $t$ is real. The area of triangle $ABC$ is $7$ square units. Determine the two possible complex numbers represented by $C$ .
(5)
(Total for Question 2 is 5 marks)
3.
The points $A$ , $B$ and $C$ are consecutive vertices of a parallelogram $ABCD$ and are represented by $1+i$ , $5+2i$ and $t+5i$ , where $t>5$ . The diagonals of the parallelogram are perpendicular. Find the value of $t$ and the complex number represented by $D$ .
(6)
(Total for Question 3 is 6 marks)
4.
The points $A$ , $B$ and $C$ are represented by $1+i$ , $7+i$ and $3+5i$ , respectively. Find the complex number represented by the point that is equidistant from $A$ , $B$ and $C$ , and find this exact common distance.
(6)
(Total for Question 4 is 6 marks)
5.
The points $A$ , $B$ and $C$ are represented by $-1+2i$ , $2-i$ and $6+i$ , respectively. Find the exact value of $\cos ABC$ , state whether angle $ABC$ is acute or obtuse, and determine the exact area of triangle $ABC$ .
(6)
(Total for Question 5 is 6 marks)

CP-2.5 · Convert between the Cartesian form and the modulus-argument form of a complex number.

Explanation

For $z=x+iy$ , the modulus is $r=|z|=\sqrt{x^2+y^2}$ and an argument $\theta$ is the directed angle from the positive real axis to the vector representing $z$ . Thus $z=r(\cos\theta+i\sin\theta)$ , or $z=re^{i\theta}$ .
When converting from Cartesian form, calculate $r$ and use $\tan\theta=y/x$ only as a reference angle: the signs of $x$ and $y$ determine the correct quadrant.
The principal argument lies in $(-\pi,\pi]$ .
To return to Cartesian form, use $x=r\cos\theta$ and $y=r\sin\theta$ .
Exact special-angle values should remain exact, and radians are used throughout this topic.

Worked example

Express $-\sqrt3+i$ in modulus-argument form using its principal argument.

1. $r=\sqrt{(-\sqrt3)^2+1^2}=2$ .
2.The point lies in the second quadrant and has reference angle $\pi/6$ .
3.Its principal argument is $\pi-\pi/6=5\pi/6$ .

Answer: $2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)$ , equivalently $2e^{5\pi i/6}$ .

Common mistakes

Don't fall into the trap of using $\tan^{-1}(y/x)$ as the final argument without checking the quadrant.
Don't fall into the trap of writing the modulus as $x^2+y^2$ instead of $\sqrt{x^2+y^2}$ .
Don't fall into the trap of giving a degree angle when the complex-number question requires radians.

Exam tip

State the quadrant before choosing an argument, especially when the real part is negative.

Tier 1 · Easy

1.
Express $1+\sqrt3i$ in the form $r(\cos\theta+i\sin\theta)$ , where $r>0$ and $-\pi<\theta\leq\pi$ .
(2)
(Total for Question 1 is 2 marks)
2.
A complex number $z=x+iy$ has modulus $6$ , satisfies $y=-\sqrt3x$ , and has negative real part. Determine $z$ and its principal argument.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Express $-2-2i$ in modulus-argument form using its principal argument.
(3)
(Total for Question 1 is 3 marks)
2.
A complex number $z$ has modulus $5$ and imaginary part $-4$ . Find the two possible values of $z$ in Cartesian form and give each principal argument exactly.
(4)
(Total for Question 2 is 4 marks)
3.
A complex number $z$ has modulus $51$ and has negative real and imaginary parts. The ratio $|\operatorname{Im}(z)|:|\operatorname{Re}(z)|$ is $8:15$ . Determine $z$ in Cartesian form and give its principal argument.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
The complex number $z=a+ib$ has modulus $10$ , positive real part and principal argument $-\tan^{-1}(3/4)$ . Determine $z$ in Cartesian form.
(4)
(Total for Question 1 is 4 marks)
2.
The complex number $z$ has modulus $25$ , and its principal argument is twice the principal argument of $3+4i$ . Determine $z$ in Cartesian form.
(5)
(Total for Question 2 is 5 marks)
3.
Let $z=x+iy$ , where $x,y\in\mathbb R$ . Given that $|z|=8$ , $x+y=4\sqrt2$ and $x<0$ , determine $z$ in Cartesian form and find its principal argument exactly.
(6)
(Total for Question 3 is 6 marks)

CP-2.6 · Multiply and divide complex numbers in modulus-argument form.

Explanation

If $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$ , then $z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}$ : moduli multiply and arguments add. For division, $\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}e^{i(\theta_1-\theta_2)}$ , provided $z_2\ne0$ : moduli divide and arguments subtract.
These results also follow from the compound-angle formulae in trigonometric form.
Any argument may be adjusted by an integer multiple of $2\pi$ ; if a principal argument is requested, place it in $(-\pi,\pi]$ .
Geometrically, multiplication combines a scaling by the modulus with a rotation by the argument, which helps check whether the resulting modulus and quadrant are sensible.
Final form should retain a positive modulus.

Worked example

Let $z_1=3e^{5\pi i/6}$ and $z_2=6e^{-2\pi i/3}$ . Find $z_1/z_2$ with principal argument.

1.Divide the moduli: $3/6=1/2$ .
2.Subtract the arguments: $5\pi/6-(-2\pi/3)=3\pi/2$ .
3.Subtract $2\pi$ to obtain the principal argument $-\pi/2$ .

Answer: $\dfrac12e^{-\pi i/2}$ .

Common mistakes

Don't fall into the trap of adding the moduli when multiplying complex numbers in modulus-argument form.
Don't fall into the trap of subtracting the numerator argument from the denominator argument during division.
Don't fall into the trap of leaving an argument such as $3\pi/2$ when the principal argument has been requested.

Exam tip

Write separate modulus and argument lines before combining them, then normalise the argument only at the end.

Tier 1 · Easy

1.
Express $2e^{i\pi/6}\times3e^{i\pi/4}$ in the form $re^{i\theta}$ .
(2)
(Total for Question 1 is 2 marks)
2.
$z_1$ has modulus $3$ and argument $2\pi/5$ ; $z_2$ has modulus $4$ and argument $-3\pi/5$ . Find $|z_1z_2|$ and $\arg(z_1z_2)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Express $\dfrac{12e^{5\pi i/6}}{3e^{-\pi i/4}}$ in the form $re^{i\theta}$ using the principal argument.
(3)
(Total for Question 1 is 3 marks)
2.
Evaluate $\dfrac{(2e^{i\pi/3})(3e^{5\pi i/6})}{\sqrt3e^{-i\pi/2}}$ , giving the result in Cartesian form.
(4)
(Total for Question 2 is 4 marks)
3.
Let $z=\sqrt3e^{-5\pi i/8}$ and $w=3e^{\pi i/12}$ . Find the modulus and principal argument of $\dfrac{z^4}{w}$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Given that $z(2e^{i\pi/7})^3=16e^{-2\pi i/7}$ , determine $z$ in the form $re^{i\theta}$ , where $-\pi<\theta\leq\pi$ .
(4)
(Total for Question 1 is 4 marks)
2.
Complex numbers $z_1$ and $z_2$ satisfy $z_1z_2=12e^{5\pi i/6}$ and $z_1/z_2=3e^{-\pi i/2}$ . Given that $\operatorname{Re}(z_1)>0$ , determine $z_1$ and $z_2$ in the form $re^{i\theta}$ , where $r>0$ and $-\pi<\theta\leq\pi$ .
(6)
(Total for Question 2 is 6 marks)
3.
Let $z=3e^{\pi i/8}$ and $w=\dfrac{z^2}{\overline z}$ . Express $w$ first with its principal argument and then in exact Cartesian form.
(5)
(Total for Question 3 is 5 marks)
4.
Non-zero complex numbers $z$ and $w$ satisfy $z^2w=32e^{i\pi/12}$ and $zw^2=16e^{-i\pi/3}$ . Given that $0<\arg z<\pi/3$ and $-\pi/2<\arg w<0$ , determine $z$ and $w$ in modulus-argument form using principal arguments.
(6)
(Total for Question 4 is 6 marks)
5.
A non-zero complex number $z$ satisfies $z^3\overline z^2=32e^{-i\pi/3}$ . Determine $z$ in modulus-argument form with principal argument, and hence give $z$ in exact Cartesian form.
(6)
(Total for Question 5 is 6 marks)

CP-2.7 · Construct and interpret simple loci in the Argand diagram such as |z - a| > r and arg(z - a) = theta.

Explanation

A locus translates complex notation into distance and direction. The equation $|z-a|=r$ is a circle of radius $r$ centred at the point representing $a$ ; $|z-a|\le r$ selects its closed interior and $|z-a|>r$ its exterior without the boundary.
The equation $|z-a|=|z-b|$ is the perpendicular bisector of the segment joining the points representing $a$ and $b$ , while $|z-a|\le|z-b|$ selects the side at least as close to the point representing $a$ .
The condition $\arg(z-a)=\beta$ is a ray from that point at directed angle $\beta$ , excluding the endpoint.
Double argument inequalities select a sector.
Sketch boundaries first, then use solid or dashed curves to show inclusion correctly.

A circle locus and an argument-ray locus measured from the same point representing

a

Worked example

Let $z=x+iy$ . Describe the region $|z-1|\le|z-(5+2i)|$ in Cartesian form.

1.Square both non-negative distances: $(x-1)^2+y^2\le(x-5)^2+(y-2)^2$ .
2.Expand and cancel $x^2$ and $y^2$ .
3. $-2x+1\le-10x+29-4y$ , so $2x+y\le7$ .
4.The boundary is the perpendicular bisector, and the inequality selects the side containing the point representing $1$ .

Answer: The closed half-plane $2x+y\le7$ .

Common mistakes

Don't centre the circle $|z-a|=r$ at $-a$ instead of at $a$ .
Don't fall into the trap of including the endpoint $z=a$ in an argument locus even though $\arg 0$ is undefined.
Don't fall into the trap of drawing a full line for $\arg(z-a)=\beta$ instead of the directed ray from $a$ .

Exam tip

For a locus sketch, label the centre or endpoint, state whether each boundary is included, and shade the side selected by a test point.

Tier 1 · Easy

1.
Describe geometrically the locus $|z-(2-i)|>3$ , stating whether its boundary is included.
(2)
(Total for Question 1 is 2 marks)
2.
Let $z=x+iy$ . Find the complex number $z$ satisfying both $|z-(1+i)|=|z-(5-i)|$ and $\operatorname{Im}(z)=2$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Let $z=x+iy$ . Find a Cartesian description of the locus $\arg(z+1)=\pi/4$ , including the required restriction on $x$ .
(3)
(Total for Question 1 is 3 marks)
2.
A region satisfies $1\leq|z-(2+i)|\leq3$ . Describe the region, including its boundaries, and find its exact area.
(4)
(Total for Question 2 is 4 marks)
3.
Let $z=x+iy$ . Describe in Cartesian form the region satisfying both $|z-(2+i)|\leq|z-(-2-5i)|$ and $|z-(2-i)|<3$ . State whether each boundary is included, and determine whether the origin lies in the region.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
A region is defined by $|z-(1-i)|<4$ and $0<\arg(z-(1-i))<\pi/3$ . Sketch the region and determine its exact area.
(5)
(Total for Question 1 is 5 marks)
2.
A region satisfies $|z|\leq5$ and $|z-5|\leq|z|$ . Describe the region with its boundaries and determine its exact area.
(6)
(Total for Question 2 is 6 marks)
3.
Find the complex number $z$ satisfying $|z-(2+i)|=|z-(-2+5i)|$ and $\arg(z-(1-i))=3\pi/4$ . Show that the point found lies on the required argument ray, and state whether the endpoint of the ray is included.
(6)
(Total for Question 3 is 6 marks)
4.
The locus $|z-(4+3i)|=3$ is a circle. Two rays from the origin are tangent to this circle. Determine the complex number at each point of contact and find the exact argument of each ray.
(6)
(Total for Question 4 is 6 marks)
5.
A region satisfies both $|z|\leq2$ and $|z-(1+i)|\leq\sqrt2$ . Describe the region, including its boundaries, and find its exact area.
(7)
(Total for Question 5 is 7 marks)

CP-2.8 · Understand de Moivre's theorem and use it to find multiple angle formulae and sums of series.

Explanation

De Moivre's theorem states $[\cos\theta+i\sin\theta]^n=\cos(n\theta)+i\sin(n\theta)$ for every integer $n$ .
Expanding the left side and equating real or imaginary parts produces multiple-angle formulae for $\cos p\theta$ and $\sin q\theta$ in powers of $\cos\theta$ and $\sin\theta$ .
The relations $z+z^{-1}=2\cos\theta$ and $z-z^{-1}=2i\sin\theta$ , with $z=e^{i\theta}$ , also convert powers back into multiple angles and support formulae involving $\tan r\theta$ .
Finite trigonometric sums are handled by summing a complex geometric series and then taking its real or imaginary part.
The chosen part and any non-zero denominator condition must be stated.

Worked example

Use de Moivre's theorem to express $\cos4\theta$ in powers of $\cos\theta$ .

1.Expand $(\cos\theta+i\sin\theta)^4$ and equate real parts.
2. $\cos4\theta=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta$ .
3.Substitute $\sin^2\theta=1-\cos^2\theta$ throughout.
4.Collecting powers gives $\cos4\theta=8\cos^4\theta-8\cos^2\theta+1$ .

Answer: $\cos4\theta=8\cos^4\theta-8\cos^2\theta+1$ .

Common mistakes

Don't fall into the trap of equating the complete binomial expansion to only $\cos n\theta$ instead of taking its real part.
Don't fall into the trap of dropping the powers of $i$ before replacing $i^2$ by $-1$ .
Don't fall into the trap of taking the real part of a geometric series when the required sum contains sine terms.

Exam tip

State whether real or imaginary parts are being equated before simplifying a de Moivre expansion.

Tier 1 · Easy

1.
Use de Moivre's theorem to write $(\cos\theta+i\sin\theta)^5$ in modulus-argument form.
(1)
(Total for Question 1 is 1 mark)
2.
Use de Moivre's theorem to show that $(1+i)^8=16$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Starting from de Moivre's theorem, prove the identity $\sin3\theta=3\sin\theta-4\sin^3\theta$ .
(4)
(Total for Question 1 is 4 marks)
2.
Use de Moivre's theorem to prove the identity $\tan3\theta=\dfrac{3t-t^3}{1-3t^2}$ , where $t=\tan\theta$ .
(5)
(Total for Question 2 is 5 marks)
3.
Use the expansion $\displaystyle\sin^5\theta=\frac1{16}(\sin5\theta-5\sin3\theta+10\sin\theta)$ to find the exact value of $\displaystyle\int_0^\pi\sin^5\theta\,d\theta$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Using a complex geometric series, evaluate exactly $S=\displaystyle\sum_{r=0}^{5}\cos\left(\dfrac{r\pi}{4}\right)$ .
(5)
(Total for Question 1 is 5 marks)
2.
Show that $\displaystyle\sum_{r=1}^{n}\sin r\theta=\dfrac{\sin(n\theta/2)\sin((n+1)\theta/2)}{\sin(\theta/2)}$ , for $\theta\ne2k\pi$ , where $k\in\mathbb Z$ .
(6)
(Total for Question 2 is 6 marks)
3.
Using a complex geometric series, evaluate exactly $\displaystyle\sum_{r=1}^{7}2^r\cos\left(\frac{r\pi}{6}\right)$ .
(6)
(Total for Question 3 is 6 marks)
4.
For $\theta\ne2k\pi$ , where $k\in\mathbb Z$ , use a complex geometric series to show that $\displaystyle\sum_{r=1}^{n}r\cos(r\theta)=\dfrac{(n+1)\cos(n\theta)-n\cos((n+1)\theta)-1}{2(1-\cos\theta)}$ . Hence evaluate $\displaystyle\sum_{r=1}^{8}r\cos(2r\pi/3)$ exactly.
(8)
(Total for Question 4 is 8 marks)
5.
You may use $\tan4\theta=\dfrac{4t(1-t^2)}{1-6t^2+t^4}$ , where $t=\tan\theta$ and both sides are defined. Show that $\tan(\pi/16)$ , $\tan(5\pi/16)$ , $\tan(9\pi/16)$ and $\tan(13\pi/16)$ are the roots of $t^4+4t^3-6t^2-4t+1=0$ . Hence write down their product.
(6)
(Total for Question 5 is 6 marks)

CP-2.9 · Know and use the definition e^(i theta) = cos theta + i sin theta and the form z = r e^(i theta).

Explanation

Euler's definition $e^{i\theta}=\cos\theta+i\sin\theta$ identifies a unit complex number with argument $\theta$ . Hence $z=re^{i\theta}$ has modulus $r$ and argument $\theta$ , and arguments differing by $2\pi k$ represent the same number.
Replacing $\theta$ by $-\theta$ gives $e^{-i\theta}=\cos\theta-i\sin\theta$ . Adding and subtracting these equations yields $\cos\theta=\dfrac12(e^{i\theta}+e^{-i\theta})$ and $\sin\theta=\dfrac1{2i}(e^{i\theta}-e^{-i\theta})$ .
Exponential form turns multiplication into addition of arguments and often simplifies symmetric trigonometric expressions.
The trigonometric and exponential forms are interchangeable, so a final answer should use the form requested.
It must not be treated as a real exponential: $|e^{i\theta}|=1$ for every real $\theta$ .

Worked example

Use Euler's definition to simplify $e^{3i\theta}-e^{-3i\theta}$ .

1. $e^{3i\theta}=\cos3\theta+i\sin3\theta$ .
2. $e^{-3i\theta}=\cos3\theta-i\sin3\theta$ .
3.Subtracting cancels the cosine terms and doubles the imaginary sine term.

Answer: $e^{3i\theta}-e^{-3i\theta}=2i\sin3\theta$ .

Common mistakes

Don't fall into the trap of writing $e^{-i\theta}=-\cos\theta-i\sin\theta$ instead of $\cos\theta-i\sin\theta$ .
Don't fall into the trap of omitting the factor $i$ from $e^{i\theta}-e^{-i\theta}=2i\sin\theta$ .
Don't fall into the trap of assuming $e^{i\theta}$ has real-exponential growth instead of modulus $1$ .

Exam tip

Pair exponentials with opposite arguments, then use their sum for cosine or their difference for sine.

Tier 1 · Easy

1.
Express $4\left(\cos\dfrac{2\pi}{5}+i\sin\dfrac{2\pi}{5}\right)$ in exponential form.
(1)
(Total for Question 1 is 1 mark)
2.
The complex number $z$ is given by $z=7e^{-i\pi/6}$ . Express $z$ in Cartesian form, and state its real and imaginary parts.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Using Euler's definition, show that $e^{2i\theta}+e^{-2i\theta}=2\cos2\theta$ .
(3)
(Total for Question 1 is 3 marks)
2.
Solve $e^{2i\theta}-e^{i\theta}+1=0$ for $0\leq\theta<2\pi$ .
(4)
(Total for Question 2 is 4 marks)
3.
Using $\displaystyle\cos\alpha=\frac{e^{i\alpha}+e^{-i\alpha}}2$ , prove that $2\cos5\theta\cos4\theta=\cos9\theta+\cos\theta$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
For $-\pi<\theta<\pi$ , prove that $1+e^{i\theta}=2\cos(\theta/2)e^{i\theta/2}$ .
(4)
(Total for Question 1 is 4 marks)
2.
The complex number $z$ satisfies $ze^{i\pi/4}=\overline z$ and $|z|=3$ . Find the possible values of $z$ in the form $re^{i\theta}$ , where $r>0$ and $-\pi<\theta\leq\pi$ .
(5)
(Total for Question 2 is 5 marks)
3.
For $0\leq\theta<2\pi$ , let $w=\dfrac{1+3e^{i\theta}}{3+e^{i\theta}}$ . Find $\operatorname{Re}(w)$ and $\operatorname{Im}(w)$ in terms of $\theta$ . Hence find the values of $\theta$ for which $w$ is purely imaginary and state the corresponding value of $w$ for each value of $\theta$ .
(6)
(Total for Question 3 is 6 marks)
4.
For $-\pi<\theta<\pi$ , use Euler's definition to prove that $\dfrac{1-e^{i\theta}}{1+e^{i\theta}}=-i\tan(\theta/2)$ . Hence solve $\dfrac{1-e^{i\theta}}{1+e^{i\theta}}=-\dfrac{i}{\sqrt3}$ in this interval.
(6)
(Total for Question 4 is 6 marks)
5.
For $0\leq\theta<2\pi$ , the complex number $w$ is defined by $w=2e^{i\theta}+e^{-i\theta}$ . Writing $w=x+iy$ , find a Cartesian equation for the curve traced by $w$ and find the coordinates of its intercepts with the real and imaginary axes.
(6)
(Total for Question 5 is 6 marks)

CP-2.10 · Find the n distinct nth roots of r e^(i theta) for r != 0 and know that they form the vertices of a regular n-gon in the Argand diagram.

Explanation

To solve $z^n=re^{i\theta}$ with $r>0$ , allow every argument of the right side: $\theta+2k\pi$ . Taking nth roots gives $z=r^{1/n}e^{i(\theta+2k\pi)/n}$ for $k=0,1,\ldots,n-1$ .
These are the $n$ distinct roots; larger integer values repeat them.
All roots have modulus $r^{1/n}$ and consecutive arguments differ by $2\pi/n$ , so their points are equally spaced on a circle centred at the origin and form a regular $n$ -gon.
The polygon may be rotated, depending on $\theta$ , but its size and angular spacing are fixed.
Each proposed root can be checked by raising it to the nth power and reducing its argument modulo $2\pi$ .

Ten distinct tenth roots equally spaced on their common modulus circle.

Worked example

Find all solutions of $z^3=8e^{i\pi}$ in exponential form.

1.Each root has modulus $8^{1/3}=2$ .
2.The arguments are $\dfrac{\pi+2k\pi}{3}$ for $k=0,1,2$ .
3.These arguments are $\pi/3$ , $\pi$ and $5\pi/3$ .

Answer: $z=2e^{i\pi/3}$ , $z=2e^{i\pi}$ or $z=2e^{5\pi i/3}$ .

Common mistakes

Don't fall into the trap of using only the principal argument $\theta$ and therefore finding only one nth root.
Don't fall into the trap of adding $2k\pi$ after dividing by $n$ instead of before dividing.
Don't fall into the trap of giving $n+1$ values by including both endpoints of one complete argument cycle.

Exam tip

Use $k=0$ to $n-1$ explicitly and check that consecutive root arguments differ by exactly $2\pi/n$ .

Tier 1 · Easy

1.
Find the three cube roots of $8$ , giving them in exponential form.
(3)
(Total for Question 1 is 3 marks)
2.
One of the fourth roots of a non-zero complex number $w$ is $\sqrt3+i$ . Determine $w$ in Cartesian form.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Find the four roots of $z^4=16e^{i\pi/3}$ , giving their arguments in the interval $0\leq\theta<2\pi$ .
(4)
(Total for Question 1 is 4 marks)
2.
Two adjacent vertices of a regular $n$ -gon centred at the origin are represented by the complex numbers $2+2i$ and $2\sqrt2i$ . Determine $n$ and the complex number represented by the vertex adjacent to $2\sqrt2i$ on the opposite side from $2+2i$ .
(5)
(Total for Question 2 is 5 marks)
3.
The six roots of $z^6=64e^{i\pi/3}$ are plotted on an Argand diagram. Find the roots whose principal arguments satisfy $|\arg z|<\pi/3$ , and determine their product in exact modulus-argument form.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The roots of $z^5=32e^{-i\pi/2}$ are plotted on an Argand diagram. Find all five roots in exponential form and determine the exact area of the regular pentagon they form.
(6)
(Total for Question 1 is 6 marks)
2.
Two adjacent roots of $z^8=w$ are $u$ and $v$ . They satisfy $|u|=|v|=2$ and $u+v=2\sqrt{2+\sqrt2}$ , a positive real number. Determine $w$ and give all eight roots in exponential form.
(6)
(Total for Question 2 is 6 marks)
3.
The eight roots of $z^8=256i$ are plotted on an Argand diagram. Without listing all eight roots separately, find the sum of the roots with positive real part, giving the answer in exact modulus-argument form.
(6)
(Total for Question 3 is 6 marks)
4.
The six roots of $z^6=729i$ are plotted on an Argand diagram. Find the root with the greatest real part, giving it in exponential form and exact Cartesian form. Hence find the exact perimeter of the regular polygon formed by all six roots.
(6)
(Total for Question 4 is 6 marks)
5.
The roots of $z^n=w$ , where $n\geq3$ , form a regular polygon centred at the origin. Each root has modulus $5$ , adjacent roots are distance $5\sqrt{2-\sqrt2}$ apart, and one root is $5e^{i\pi/20}$ . Determine $n$ , find $w$ in exact Cartesian form, and find the product of the roots with positive imaginary part in exact modulus-argument form.
(8)
(Total for Question 5 is 8 marks)

CP-2.11 · Use complex roots of unity to solve geometric problems.

Explanation

The nth roots of unity solve $z^n=1$ and are $1,\omega,\omega^2,\ldots,\omega^{n-1}$ for $\omega=e^{2\pi i/n}$ . They lie on the unit circle as the vertices of a regular $n$ -gon.
For a non-real root $\omega\ne1$ , factorising $\omega^n-1$ gives $1+\omega+\cdots+\omega^{n-1}=0$ .
Complex methods encode geometry efficiently: the distance between the points representing $a$ and $b$ is $|a-b|$ , and multiplication by $\omega$ rotates every point representing a complex number through $2\pi/n$ without changing lengths.
Algebraic identities among powers of $\omega$ can therefore prove equal sides, angles, parallelism and ratios.
Any cancellation involving $\omega-1$ requires the explicit fact that $\omega\ne1$ .

Worked example

Let $\omega=e^{2\pi i/3}$ . Show that the points represented by $0$ , $1$ and $1+\omega$ form an equilateral triangle.

1.The side from $0$ to $1$ has length $|1|=1$ .
2.The side from $1$ to $1+\omega$ has length $|\omega|=1$ .
3.Since $1+\omega+\omega^2=0$ , $1+\omega=-\omega^2$ , so $|1+\omega|=1$ .
4.All three side lengths are $1$ .

Answer: The three points form an equilateral triangle of side length $1$ .

Common mistakes

Don't fall into the trap of using $1+\omega+\cdots+\omega^{n-1}=0$ for the root $\omega=1$ .
Don't fall into the trap of calculating a geometric distance as $a-b$ without taking the modulus.
Don't fall into the trap of claiming that multiplication by $\omega$ changes lengths even though $|\omega|=1$ .

Exam tip

Translate each geometric claim into a modulus of a difference or a rotation by a unit complex number.

Tier 1 · Easy

1.
Let $\omega=e^{2\pi i/3}$ . Find the exact distance between the points represented by $1$ and $\omega$ .
(2)
(Total for Question 1 is 2 marks)
2.
Let $\zeta=e^{i\pi/3}$ . The point represented by $2-i$ is rotated anticlockwise about the origin through $\pi/3$ . Find the complex number represented by its image exactly.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Let $\omega=e^{2\pi i/3}$ . The points $A$ , $B$ and $C$ are represented by the complex numbers $1$ , $\omega$ and $\omega^2$ , respectively. Prove that triangle $ABC$ is equilateral.
(4)
(Total for Question 1 is 4 marks)
2.
Let $\zeta=e^{i\pi/3}$ . The points $A$ , $B$ , $C$ and $D$ are represented by the complex numbers $1$ , $\zeta$ , $\zeta^2$ and $\zeta^3$ , respectively. Prove that $ABCD$ is an isosceles trapezium and find its exact area.
(6)
(Total for Question 2 is 6 marks)
3.
The vertices of a regular hexagon inscribed in the unit circle are represented by the complex numbers $1,\zeta,\zeta^2,\ldots,\zeta^5$ , where $\zeta=e^{2\pi i/6}$ . Show that the product of the distances from the vertex representing $1$ to the other five vertices is $6$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
The fifth roots of unity are the vertices of a regular pentagon. Prove that the ratio of a diagonal to a side is $\dfrac{1+\sqrt5}{2}$ .
(7)
(Total for Question 1 is 7 marks)
2.
Let $\zeta=e^{2\pi i/n}$ , where $n\geq3$ is an integer. The vertices of a regular $n$ -gon are represented by the complex numbers $1,\zeta,\ldots,\zeta^{n-1}$ . Prove that, for any point represented by $p$ , the sum of the squares of its distances from all vertices is $n(|p|^2+1)$ . Hence evaluate this sum when $n=6$ and $p=2+i$ .
(7)
(Total for Question 2 is 7 marks)
3.
Write $p=1+i$ , $q=5+3i$ and $\zeta=e^{i\pi/3}$ , where $p$ and $q$ represent points $P$ and $Q$ . Find the complex numbers represented by the two possible third vertices $R_1$ and $R_2$ of an equilateral triangle with side $PQ$ . Prove that $R_1$ and $R_2$ lie on opposite sides of the line $PQ$ .
(7)
(Total for Question 3 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

CP-2.1 · Solve any quadratic equation with real coefficients. Solve cubic or quartic equations with real coefficients.

Tier 1 · Easy

Mark scheme for CP-2.1 Tier 1 · Easy
Question	Scheme	Marks
1	$x=2+3i$ or $x=2-3i$	3
(3 marks)		3
Notes
Complete the square: $x^2-4x+13=(x-2)^2+9$ . Hence $(x-2)^2=-9$ , so $x-2=\pm3i$ and therefore $x=2\pm3i$ .
2	The discriminant is $3^2-4(2)(5)=-31$ . $x=\dfrac{-3\pm\sqrt{-31}}{4}$ . $x=-\dfrac34\pm\dfrac{\sqrt{31}}4i$ .	3
(3 marks)		3
Notes
Use the quadratic formula with $a=2$ , $b=3$ and $c=5$ . The discriminant is $9-40=-31$ , so $\sqrt{-31}=i\sqrt{31}$ . Dividing by $2a=4$ gives the exact conjugate pair $-3/4\pm(\sqrt{31}/4)i$ .

Tier 2 · Standard

Mark scheme for CP-2.1 Tier 2 · Standard
Question	Scheme	Marks
1	$x=2$ , $x=i\sqrt5$ or $x=-i\sqrt5$	4
(4 marks)		4
Notes
Group the terms: $x^3-2x^2+5x-10=x^2(x-2)+5(x-2)=(x-2)(x^2+5)$ . Thus $x=2$ or $x^2=-5$ , giving $x=\pm i\sqrt5$ .
2	Completing the square gives roots $t\pm2i$ . Their common modulus is $\sqrt{t^2+4}$ . $\sqrt{t^2+4}=5$ gives $t=\pm\sqrt{21}$ . For $t=\sqrt{21}$ , the roots are $\sqrt{21}\pm2i$ . For $t=-\sqrt{21}$ , the roots are $-\sqrt{21}\pm2i$ .	5
(5 marks)		5
Notes
Complete the square: $(z-t)^2=-4$ , so the roots are $t\pm2i$ . Either root has modulus $\sqrt{t^2+4}$ . The condition $\sqrt{t^2+4}=5$ gives $t^2=21$ , hence both real branches $t=\pm\sqrt{21}$ . Substitution gives both conjugate pairs listed.
3	The conjugate root is $-2-i\sqrt7$ . The conjugate pair gives the factor $z^2+4z+11$ . The constant term shows that the third root is $3$ . $(z^2+4z+11)(z-3)=z^3+z^2-z-33$ , so $a=1$ . The roots are $-2\pm i\sqrt7$ and $3$ .	5
(5 marks)		5
Notes
Real coefficients force the conjugate root. The pair has sum $-4$ and product $11$ , giving $z^2+4z+11$ . If the remaining root is $r$ , the constant term is $-11r=-33$ , so $r=3$ . Expansion gives $z^3+z^2-z-33$ , which fixes $a=1$ and confirms all three roots.

Tier 3 · Hard

Mark scheme for CP-2.1 Tier 3 · Hard
Question	Scheme	Marks
1	$x=\dfrac{-3+i\sqrt{15}}2$ , $x=\dfrac{-3-i\sqrt{15}}2$ , $x=\dfrac{3+i\sqrt{15}}2$ or $x=\dfrac{3-i\sqrt{15}}2$	5
(5 marks)		5
Notes
Factor using conjugate linear terms: $(x^2+3x+6)(x^2-3x+6)=x^4+(12-9)x^2+36=x^4+3x^2+36$ . Solving $x^2+3x+6=0$ gives $x=(-3\pm i\sqrt{15})/2$ , while $x^2-3x+6=0$ gives $x=(3\pm i\sqrt{15})/2$ .
2	Real coefficients give the conjugate root $1-4i$ . The known pair gives the factor $z^2-2z+17$ . Since the constant term is $68=17\cdot4$ , the other factor is $z^2+4$ . Expanding gives $(z^2-2z+17)(z^2+4)=z^4-2z^3+21z^2-8z+68$ . Comparing the $z^3$ and $z$ coefficients gives $c=-2$ and $4c=-8$ , consistently. Comparing the $z^2$ coefficient gives $d=21$ . Thus $z=1\pm4i$ or $z=\pm2i$ .	7
(7 marks)		7
Notes
The conjugate $1-4i$ is also a root, so their quadratic factor is $(z-(1+4i))(z-(1-4i))=z^2-2z+17$ . The constant term $68=17\cdot4$ and the prescribed coefficient pattern are satisfied by the second factor $z^2+4$ . Expanding gives $z^4-2z^3+21z^2-8z+68$ , so $c=-2$ , $d=21$ and the linear coefficient $4c=-8$ is consistent. The two factors give the four roots $1\pm4i$ and $\pm2i$ .
3	The conjugate root is $3-5i$ . The known conjugate pair gives the factor $z^2-6z+34$ . Write the other monic quadratic factor as $z^2+Az+B$ . The constant term gives $34B=408$ , so $B=12$ . The coefficient of $z$ gives $-6B+34A=166$ , so $A=7$ . $z^2+7z+12=(z+3)(z+4)$ , so the other roots are $-3$ and $-4$ . $(z^2-6z+34)(z^2+7z+12)=z^4+z^3+4z^2+166z+408$ . Thus the roots are $3\pm5i$ , $-3$ and $-4$ , and $u=1$ , $v=4$ .	8
(8 marks)		8
Notes
Real coefficients force the conjugate $3-5i$ , so $z^2-6z+34$ divides $f$ . Let the remaining monic factor be $z^2+Az+B$ . The constant and linear coefficients determine $B=12$ and $A=7$ independently. Factor the quotient, then expand the two quadratics to read off every remaining coefficient and check the prescribed $166z$ term.
4	Putting $u=z^2$ gives $u^2+6u+25=0$ . The two values of $u$ are $-3+4i$ and $-3-4i$ . Since $(1+2i)^2=-3+4i$ , the square roots of $-3+4i$ are $\pm(1+2i)$ . Since $(1-2i)^2=-3-4i$ , the square roots of $-3-4i$ are $\pm(1-2i)$ . Thus $z=1+2i$ , $1-2i$ , $-1+2i$ or $-1-2i$ . For $w\ne0$ , multiplying $25w^4+6w^2+1=0$ by $w^{-4}$ gives $(w^{-1})^4+6(w^{-1})^2+25=0$ . Therefore each value of $w$ is the reciprocal of one of the four values of $z$ . Hence $w=(1+2i)/5$ , $(1-2i)/5$ , $(-1+2i)/5$ or $(-1-2i)/5$ .	8
(8 marks)		8
Notes
Treat the first equation as a quadratic in $z^2$ . Recognise each resulting Cartesian number as a square and include both signs. The second polynomial is the reciprocal polynomial: after division by $w^4$ , setting $z=1/w$ reproduces the first equation, so rationalise the four reciprocals.
5	Write the polynomial as $(z^2+az+b)^2$ . Comparison of the $z^3$ coefficient gives $2a=-16$ , so $a=-8$ . Comparison of the $z^2$ coefficient gives $a^2+2b=146$ , so $b=41$ . The remaining coefficients satisfy $2ab=-656$ and $b^2=1681$ , so the polynomial is $(z^2-8z+41)^2$ . $z^2-8z+41=0$ gives $z=4+5i$ or $z=4-5i$ . Each root has multiplicity $2$ . $1/(4+5i)+1/(4-5i)=8/41$ . Counting each root twice, the required sum is $16/41$ .	8
(8 marks)		8
Notes
Compare the expansion of a squared monic quadratic with the given coefficients, including the two consistency checks. Solve the repeated quadratic and retain the doubled multiplicities. Rationalising the conjugate reciprocals gives $8/41$ for one pair, which is doubled in the quartic.

CP-2.2 · Add, subtract, multiply and divide complex numbers in the form x + iy with x and y real. Understand and use the terms 'real part' and 'imaginary part'.

Tier 1 · Easy

Mark scheme for CP-2.2 Tier 1 · Easy
Question	Scheme	Marks
1	$z=8+5i$ $\operatorname{Re}(z)=8$ $\operatorname{Im}(z)=5$	2
(2 marks)		2
Notes
Add corresponding parts: $z=(3+5)+(-2+7)i=8+5i$ . Therefore the real part is $8$ and the imaginary part is $5$ .
2	Equating parts gives $a^2-b^2=5$ and $ab=6$ . $(a^2+b^2)^2=5^2+12^2=169$ , so $a^2+b^2=13$ . Hence $a^2=9$ and $b^2=4$ ; positivity gives $a=3$ and $b=2$ .	3
(3 marks)		3
Notes
Expanding gives $(a+bi)^2=(a^2-b^2)+2abi$ , so $a^2-b^2=5$ and $2ab=12$ . Also $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=25+144=169$ . Since $a$ and $b$ are real, $a^2+b^2=13$ . Adding and subtracting with $a^2-b^2=5$ gives $a^2=9$ and $b^2=4$ , and the stated positivity selects $a=3$ , $b=2$ .

Tier 2 · Standard

Mark scheme for CP-2.2 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac75+\dfrac65i$	3
(3 marks)		3
Notes
Multiply by the conjugate $2+i$ : $\dfrac{4+i}{2-i}=\dfrac{(4+i)(2+i)}{(2-i)(2+i)}=\dfrac{7+6i}{5}=\dfrac75+\dfrac65i$ .
2	$\operatorname{Re}((2+i)z)=1$ gives $2x-y=1$ . $z/(1-i)=(x-y)/2+i(x+y)/2$ . The imaginary-part condition gives $x+y=4$ . $x=5/3$ , $y=7/3$ , so $z=5/3+7i/3$ .	4
(4 marks)		4
Notes
From $(2+i)(x+iy)=(2x-y)+i(x+2y)$ , the first condition is $2x-y=1$ . Also $z/(1-i)=(x+iy)(1+i)/2=(x-y)/2+i(x+y)/2$ , so the second condition gives $x+y=4$ . Solving the two real equations gives $x=5/3$ and $y=7/3$ .
3	Expanding the left side gives $(5x+3y)+i(-x+y)$ . Equating real parts gives $5x+3y=1$ . Equating imaginary parts gives $-x+y=3$ . Solving gives $x=-1$ and $y=2$ , so $z=-1+2i$ .	4
(4 marks)		4
Notes
Use $\overline z=x-iy$ and expand both products. Equality with $1+3i$ gives the simultaneous equations $5x+3y=1$ and $-x+y=3$ . Substitution gives $x=-1$ , $y=2$ , and direct replacement in the original equation confirms the Cartesian value.

Tier 3 · Hard

Mark scheme for CP-2.2 Tier 3 · Hard
Question	Scheme	Marks
1	$\dfrac{37}{10}-\dfrac{29}{10}i$	5
(5 marks)		5
Notes
First expand the numerator: $(2+3i)(1-4i)=2-8i+3i-12i^2=14-5i$ . Then multiply by the conjugate of the denominator: $\dfrac{14-5i}{3+i}=\dfrac{(14-5i)(3-i)}{10}=\dfrac{37-29i}{10}$ .
2	$z=4+3i-(1-i)w$ . Substitution into the second equation gives $i(4+3i)+w=-4+7i$ . $i(4+3i)=-3+4i$ . Hence $w=-1+3i$ . Back-substitution gives $z=2-i$ .	5
(5 marks)		5
Notes
From the first equation, $z=4+3i-(1-i)w$ . Substitute this into the second. Since $-i(1-i)+(2+i)=1$ , the left side becomes $i(4+3i)+w=(-3+4i)+w$ . Thus $w=(-4+7i)-(-3+4i)=-1+3i$ . Then $(1-i)w=2+4i$ , so $z=4+3i-(2+4i)=2-i$ . Direct substitution checks both original complex equations.
3	$(1+2i)^2=-3+4i$ . $(1+2i)^3=(-3+4i)(1+2i)=-11-2i$ . $(3+2i)(1-i)=5-i$ . $\displaystyle\frac{-11-2i}{5-i}=\frac{(-11-2i)(5+i)}{26}$ . Therefore the Cartesian form is $-\dfrac{53}{26}-\dfrac{21}{26}i$ .	5
(5 marks)		5
Notes
Build the numerator by squaring and then multiplying once more. The denominator factors give $5-i$ , whose squared modulus is $26$ . Rationalising with $5+i$ gives numerator $-53-21i$ and denominator $26$ .

CP-2.3 · Understand and use the complex conjugate. Know that non-real roots of polynomial equations with real coefficients occur in conjugate pairs.

Tier 1 · Easy

Mark scheme for CP-2.3 Tier 1 · Easy
Question	Scheme	Marks
1	$\overline z=5+4i$ $z\overline z=41$	2
(2 marks)		2
Notes
Reverse the sign of the imaginary part to get $\overline z=5+4i$ . Then $z\overline z=(5-4i)(5+4i)=5^2+4^2=41$ .
2	The conjugate pair is $a+5i$ and $a-5i$ . Their sum gives $2a=-6$ , so the roots are $-3\pm5i$ . Their product is $34$ , so the quadratic is $x^2+6x+34$ .	3
(3 marks)		3
Notes
The roots must be $a+5i$ and $a-5i$ . Their sum is $2a=-6$ , because the coefficient of $x$ is $6$ , so $a=-3$ . Their product is $(-3)^2+5^2=34$ , giving the monic quadratic $x^2+6x+34$ .

Tier 2 · Standard

Mark scheme for CP-2.3 Tier 2 · Standard
Question	Scheme	Marks
1	$x=2+i$ , $x=2-i$ or $x=-3$	4
(4 marks)		4
Notes
Because the coefficients are real, $2-i$ is also a root. Their factor is $(x-2-i)(x-2+i)=(x-2)^2+1=x^2-4x+5$ . Division or comparison gives $p(x)=(x^2-4x+5)(x+3)$ , so the third root is $-3$ .
2	The conjugate root is $4-3i$ . The conjugate pair gives the factor $x^2-8x+25$ . The product of the roots is $-50$ , so $25r=-50$ gives the third root $r=-2$ . The factors are $(x^2-8x+25)(x+2)$ . The polynomial is $x^3-6x^2+9x+50$ , so $a=-6$ and $b=9$ .	5
(5 marks)		5
Notes
The conjugate root is $4-3i$ , and the pair gives $(x-(4+3i))(x-(4-3i))=x^2-8x+25$ . For a monic cubic, the constant term is minus the product of the roots. Thus $-25r=50$ , equivalently $25r=-50$ , so the third root is $r=-2$ . Therefore the polynomial is $(x^2-8x+25)(x+2)=x^3-6x^2+9x+50$ , giving $a=-6$ and $b=9$ .
3	Because the coefficients are real, $5+2i$ is also a root. The sum of the roots is $10$ , so $-m=10$ and $m=-10$ . The product is $(5-2i)(5+2i)=29$ , so $n=29$ .	3
(3 marks)		3
Notes
A non-real root of a polynomial with real coefficients brings its conjugate. Vieta's formulae then determine both unknown coefficients: the coefficient of $z$ is the negative root sum and the constant is the root product.

Tier 3 · Hard

Mark scheme for CP-2.3 Tier 3 · Hard
Question	Scheme	Marks
1	The root $5+i$ forces the conjugate root $5-i$ . The root $-1+3i$ forces the conjugate root $-1-3i$ . The pair $5\pm i$ gives the factor $x^2-10x+26$ . The pair $-1\pm3i$ gives the factor $x^2+2x+10$ . The monic polynomial is $(x^2-10x+26)(x^2+2x+10)$ . In expanded form, it is $x^4-8x^3+16x^2-48x+260$ .	6
(6 marks)		6
Notes
Because the coefficients are real, the other roots are $5-i$ and $-1-3i$ . The first conjugate pair gives $(x-5)^2+1=x^2-10x+26$ , and the second gives $(x+1)^2+9=x^2+2x+10$ . Therefore the monic polynomial is $(x^2-10x+26)(x^2+2x+10)=x^4-8x^3+16x^2-48x+260$ .
2	$(a+b)^2=a^2+b^2+2ab$ gives $ab=-2$ . $a$ and $b$ solve $t^2+t-2=0$ , so they are $1$ and $-2$ . The condition $a>0$ fixes $a=1$ and $b=-2$ . The pair $1\pm i$ gives the factor $x^2-2x+2$ . The pair $-2\pm4i$ gives the factor $x^2+4x+20$ . Thus the quartic is $(x^2-2x+2)(x^2+4x+20)$ and its roots are $1\pm i$ and $-2\pm4i$ .	6
(6 marks)		6
Notes
From $(a+b)^2=a^2+b^2+2ab$ , $1=5+2ab$ , so $ab=-2$ . Thus $a$ and $b$ are the roots of $t^2+t-2=0$ , namely $1$ and $-2$ . The condition $a>0$ fixes $a=1$ and $b=-2$ . Real coefficients force the conjugates $1-i$ and $-2-4i$ . Pairing conjugates gives $(x-(1+i))(x-(1-i))=x^2-2x+2$ and $(x-(-2+4i))(x-(-2-4i))=x^2+4x+20$ . Hence $p=-2$ , $q=2$ , $r=4$ and $s=20$ , with roots $1\pm i$ and $-2\pm4i$ .
3	The conjugate factors are $x^2-2ax+a^2+4$ and $x^2-2bx+b^2+25$ . The coefficient of $x^3$ gives $-2(a+b)=-6$ , so $a+b=3$ . The coefficient of $x^2$ is $(a+b)^2+2ab+29$ , so $18=38+2ab$ . Hence $ab=-10$ . So $a$ and $b$ are the roots of $t^2-3t-10=0$ , namely $-2$ and $5$ . The condition $a-2b<0$ selects $a=-2$ and $b=5$ . The quartic is $(x^2+4x+8)(x^2-10x+50)=x^4-6x^3+18x^2+120x+400$ , with roots $-2\pm2i$ and $5\pm5i$ .	7
(7 marks)		7
Notes
Form the two real quadratics generated by the conjugate pairs. The cubic coefficient gives their real-part sum, while the quadratic coefficient gives their product after rewriting $a^2+b^2+4ab+29$ as $(a+b)^2+2ab+29$ . Solve the resulting quadratic and apply the non-symmetric condition to attach the correct imaginary part to each real part before expanding.

CP-2.4 · Use and interpret Argand diagrams.

Tier 1 · Easy

Mark scheme for CP-2.4 Tier 1 · Easy
Question	Scheme	Marks
1	$P=(-2,3)$ Second quadrant	2
(2 marks)		2
Notes
The real part gives the horizontal coordinate and the imaginary part gives the vertical coordinate, so $P=(-2,3)$ . Negative real part and positive imaginary part place it in the second quadrant.
2	$d=a+c-b$ for consecutive parallelogram vertices. $d=(1+2i)+(5+4i)-(4+i)=2+5i$ .	2
(2 marks)		2
Notes
For consecutive vertices, $a+c=b+d$ , so $d=a+c-b=(1+2i)+(5+4i)-(4+i)=2+5i$ . In coordinates, $A(1,2)$ , $B(4,1)$ , $C(5,4)$ and $D(2,5)$ ; both $\overrightarrow{AB}$ and $\overrightarrow{DC}$ are $(3,-1)$ .

Tier 2 · Standard

Mark scheme for CP-2.4 Tier 2 · Standard
Question	Scheme	Marks
1	$6$ square units	3
(3 marks)		3
Notes
The coordinates are $A(1,1)$ , $B(5,1)$ and $C(1,4)$ . The triangle is right-angled at $A$ , with $AB=4$ and $AC=3$ . Hence its area is $\tfrac12\times4\times3=6$ .
2	$a+b+c=3(3+3i)=9+9i$ . $C$ is represented by the complex number $3+6i$ . $c-a=(3+6i)-(1+i)=2+5i$ , so $\|c-a\|=\sqrt{29}$ . Since the real and imaginary parts of $c-a$ are both positive, $\arg(c-a)=\tan^{-1}(5/2)$ .	4
(4 marks)		4
Notes
Rearrange the supplied relation directly: $c=3g-a-b=9+9i-(1+i)-(5+2i)=3+6i$ . The directed displacement from $A$ to $C$ is $c-a=2+5i$ . Its real and imaginary components give $\|c-a\|=\sqrt{2^2+5^2}=\sqrt{29}$ . Both components are positive, so the principal argument is $\tan^{-1}(5/2)$ .
3	$r-p=9-6i=3(3-2i)$ . $q-r=3-2i$ , so $PR=3\sqrt{13}$ and $RQ=\sqrt{13}$ . Therefore $PR:RQ=3:1$ .	3
(3 marks)		3
Notes
Subtract the complex numbers representing the endpoints from the complex number representing $R$ . The displacement $r-p$ is three times $q-r$ , so their moduli are in the same factor-three ratio. The supplied collinearity and segment condition fix this as the internal division ratio.

Tier 3 · Hard

Mark scheme for CP-2.4 Tier 3 · Hard
Question	Scheme	Marks
1	$OA=OB=\sqrt{10}$ $OA\perp OB$ Area $=5$ square units	5
(5 marks)		5
Notes
The vectors are $\overrightarrow{OA}=(3,1)$ and $\overrightarrow{OB}=(-1,3)$ . Both have squared length $10$ , so $OA=OB=\sqrt{10}$ . Their scalar product is $3(-1)+1(3)=0$ , so they are perpendicular. The area is therefore $\tfrac12(\sqrt{10})(\sqrt{10})=5$ .
2	$AB=2\sqrt5$ . The perpendicular distance from $C$ to $AB$ is $d=\|t+3\|/\sqrt5$ . The area condition gives $\|t+3\|=7$ . Thus $t=4$ or $t=-10$ . The two possible complex numbers represented by $C$ are $4+4i$ and $-10+4i$ .	5
(5 marks)		5
Notes
The length $AB=\|4-2i\|=2\sqrt5$ . The line through $A(1,2)$ and $B(5,0)$ is $x+2y-5=0$ , so the perpendicular distance from $C(t,4)$ to $AB$ is $d=\|t+8-5\|/\sqrt5=\|t+3\|/\sqrt5$ . Hence the area is $\tfrac12\|AB\|d=\tfrac12(2\sqrt5)(\|t+3\|/\sqrt5)=\|t+3\|$ . Solving $\|t+3\|=7$ gives $t=4$ or $t=-10$ , so the two possible complex numbers represented by $C$ are $4+4i$ and $-10+4i$ .
3	$d=a+c-b=(t-4)+4i$ . The diagonal vectors are $\overrightarrow{AC}=(t-1,4)$ and $\overrightarrow{BD}=(t-9,2)$ . Perpendicularity gives $(t-1)(t-9)+8=0$ . Thus $t^2-10t+17=0$ . $t=5\pm2\sqrt2$ , and $t>5$ selects $t=5+2\sqrt2$ . Therefore $D$ is represented by $1+2\sqrt2+4i$ .	6
(6 marks)		6
Notes
For consecutive vertices, $a+c=b+d$ , giving $d=(t-4)+4i$ . The two diagonals have coordinate vectors $(t-1,4)$ and $(t-9,2)$ . Their scalar product is zero, so $(t-1)(t-9)+8=0$ , which simplifies to $t^2-10t+17=0$ . The stated inequality selects $5+2\sqrt2$ , and substitution gives $d=1+2\sqrt2+4i$ .
4	$AB$ is horizontal and has midpoint $(4,1)$ . Its perpendicular bisector is therefore $x=4$ . $AC$ has midpoint $(2,3)$ and gradient $2$ . Its perpendicular bisector has equation $y-3=-\tfrac12(x-2)$ . The two perpendicular bisectors meet at $(4,2)$ , so the required point is represented by $4+2i$ . Its distance from $A$ is $\sqrt{3^2+1^2}=\sqrt{10}$ , which is the exact common distance.	6
(6 marks)		6
Notes
Find two perpendicular bisectors in coordinates read from the Argand diagram. Their unique intersection is equidistant from the three vertices. A distance from this point to any one vertex gives the common distance, and the other two squared distances also equal $10$ .
5	The directed displacements from $B$ to $A$ and $C$ are $(-3,3)$ and $(4,2)$ . Their scalar product is $(-3)(4)+(3)(2)=-6$ . Their lengths are $3\sqrt2$ and $2\sqrt5$ , respectively. Therefore $\cos ABC=-6/(6\sqrt{10})=-1/\sqrt{10}$ . Since this cosine is negative, angle $ABC$ is obtuse. Since $\sin ABC=3/\sqrt{10}$ , the area is $\tfrac12(3\sqrt2)(2\sqrt5)(3/\sqrt{10})=9$ square units.	6
(6 marks)		6
Notes
Read the two displacement vectors from the Argand coordinates. Their scalar product divided by the product of their lengths gives the exact cosine and its sign determines the type of angle. The exact sine then follows from the cosine, and $\tfrac12ab\sin C$ gives the area.

CP-2.5 · Convert between the Cartesian form and the modulus-argument form of a complex number.

Tier 1 · Easy

Mark scheme for CP-2.5 Tier 1 · Easy
Question	Scheme	Marks
1	$2\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)$	2
(2 marks)		2
Notes
The modulus is $\sqrt{1^2+(\sqrt3)^2}=2$ . Its real and imaginary parts are both positive and $\tan\theta=\sqrt3$ , so $\theta=\pi/3$ .
2	The conditions place $z$ in quadrant II, so $\arg z=2\pi/3$ . $z=6(\cos(2\pi/3)+i\sin(2\pi/3))$ . $z=-3+3\sqrt3i$ .	3
(3 marks)		3
Notes
The line $y=-\sqrt3x$ with $x<0$ places $z$ in the second quadrant, so its principal argument is $2\pi/3$ . Therefore $z=6(\cos(2\pi/3)+i\sin(2\pi/3))=-3+3\sqrt3i$ . The other point on the line and modulus circle has positive real part, so it is excluded.

Tier 2 · Standard

Mark scheme for CP-2.5 Tier 2 · Standard
Question	Scheme	Marks
1	$2\sqrt2\left(\cos\left(-\dfrac{3\pi}{4}\right)+i\sin\left(-\dfrac{3\pi}{4}\right)\right)$	3
(3 marks)		3
Notes
The modulus is $\sqrt{(-2)^2+(-2)^2}=2\sqrt2$ . The point is in the third quadrant, whose principal argument is $-3\pi/4$ . Substitute these into $r(\cos\theta+i\sin\theta)$ .
2	$x^2+16=25$ , so $x=3$ or $x=-3$ . The Cartesian possibilities are $3-4i$ and $-3-4i$ . $\arg(3-4i)=-\tan^{-1}(4/3)$ . $\arg(-3-4i)=-\pi+\tan^{-1}(4/3)$ .	4
(4 marks)		4
Notes
Write $z=x-4i$ . From $\|z\|=5$ , $x^2+16=25$ , so both branches $x=3$ and $x=-3$ are possible. The first point is in quadrant IV, giving $-\tan^{-1}(4/3)$ . The second is in quadrant III, whose principal argument is $-\pi+\tan^{-1}(4/3)$ .
3	The component ratio $8:15$ has hypotenuse ratio $17$ . The scale factor is $51/17=3$ , so the component magnitudes are $24$ and $45$ . Hence $z=-45-24i$ . $\arg z=-\pi+\tan^{-1}(8/15)$ .	4
(4 marks)		4
Notes
The ratio $8:15$ belongs to an $8$ - $15$ - $17$ right triangle. Since the modulus is three times $17$ , the imaginary and real component magnitudes are $24$ and $45$ . Both stated signs are negative, so $z=-45-24i$ lies in quadrant III; its principal argument is $-\pi$ plus the reference angle $\tan^{-1}(8/15)$ .

Tier 3 · Hard

Mark scheme for CP-2.5 Tier 3 · Hard
Question	Scheme	Marks
1	$z=8-6i$	4
(4 marks)		4
Notes
A reference triangle for $\tan^{-1}(3/4)$ has adjacent, opposite and hypotenuse in the ratio $4:3:5$ . The negative argument with positive real part places $z$ in the fourth quadrant, so $\cos\theta=4/5$ and $\sin\theta=-3/5$ . Hence $z=10(4/5-3i/5)=8-6i$ .
2	The complex number $3+4i$ has argument $\theta=\tan^{-1}(4/3)$ , where $\pi/4<\theta<\pi/2$ . Squaring gives $(3+4i)^2=-7+24i$ . The number $-7+24i$ has modulus $25$ . Since $\pi/2<2\theta<\pi$ , its principal argument is $2\theta$ with no adjustment. Hence $z=-7+24i$ .	5
(5 marks)		5
Notes
The number $3+4i$ has argument $\theta=\tan^{-1}(4/3)$ because its real and imaginary parts are positive. Since $4/3>1$ , $\pi/4<\theta<\pi/2$ , so $\pi/2<2\theta<\pi$ . Squaring gives $(3+4i)^2=-7+24i$ , which has modulus $\sqrt{(-7)^2+24^2}=25$ and lies in quadrant II. Its principal argument is therefore exactly $2\theta$ , with no $2\pi$ adjustment. Hence the required complex number is $z=-7+24i$ .
3	$x^2+y^2=64$ and $(x+y)^2=32$ , so $xy=-16$ . $x$ and $y$ are the roots of $t^2-4\sqrt2t-16=0$ . These roots are $2\sqrt2\pm2\sqrt6$ . The condition $x<0$ gives $x=2\sqrt2-2\sqrt6$ and $y=2\sqrt2+2\sqrt6$ . Thus $z=2(\sqrt2-\sqrt6)+2(\sqrt2+\sqrt6)i$ . Its principal argument is $7\pi/12$ .	6
(6 marks)		6
Notes
Squaring the sum gives $32=x^2+y^2+2xy=64+2xy$ , so $xy=-16$ . The components are therefore the roots of $t^2-4\sqrt2t-16=0$ , namely $2\sqrt2\pm2\sqrt6$ . The sign condition assigns the smaller root to $x$ . The resulting components equal $8\cos(7\pi/12)$ and $8\sin(7\pi/12)$ , so the point is in quadrant II with principal argument $7\pi/12$ .

CP-2.6 · Multiply and divide complex numbers in modulus-argument form.

Tier 1 · Easy

Mark scheme for CP-2.6 Tier 1 · Easy
Question	Scheme	Marks
1	$6e^{5\pi i/12}$	2
(2 marks)		2
Notes
Multiply the moduli to get $2\times3=6$ and add the arguments: $\pi/6+\pi/4=2\pi/12+3\pi/12=5\pi/12$ . Therefore the product is $6e^{5\pi i/12}$ .
2	$\|z_1z_2\|=3\times4=12$ . $\arg(z_1z_2)=2\pi/5-3\pi/5=-\pi/5$ .	2
(2 marks)		2
Notes
For a product, multiply the moduli and add the arguments. This gives modulus $3\cdot4=12$ and argument $2\pi/5+(-3\pi/5)=-\pi/5$ . Since $-\pi<-\pi/5\leq\pi$ , no further $2\pi$ adjustment is needed for the principal argument.

Tier 2 · Standard

Mark scheme for CP-2.6 Tier 2 · Standard
Question	Scheme	Marks
1	$4e^{-11\pi i/12}$	3
(3 marks)		3
Notes
Divide the moduli to get $4$ . Subtract the arguments: $5\pi/6-(-\pi/4)=13\pi/12$ . This exceeds $\pi$ , so subtract $2\pi$ to obtain the principal argument $-11\pi/12$ .
2	The modulus is $2\sqrt3$ . The unreduced argument is $5\pi/3$ . The principal exponential form is $2\sqrt3e^{-i\pi/3}$ . The Cartesian result is $\sqrt3-3i$ .	4
(4 marks)		4
Notes
The modulus is $6/\sqrt3=2\sqrt3$ . The argument is $\pi/3+5\pi/6-(-\pi/2)=5\pi/3$ , whose principal equivalent is $-\pi/3$ . Hence the value is $2\sqrt3(\cos(-\pi/3)+i\sin(-\pi/3))=2\sqrt3(\tfrac12-\tfrac{\sqrt3}{2}i)=\sqrt3-3i$ .
3	$\|z^4/w\|=(\sqrt3)^4/3=3$ . An argument of $z^4/w$ is $4(-5\pi/8)-\pi/12=-31\pi/12$ . Adding $2\pi$ gives the equivalent argument $-7\pi/12$ . Since $-7\pi/12\in(-\pi,\pi]$ , the modulus is $3$ and the principal argument is $-7\pi/12$ .	4
(4 marks)		4
Notes
Raising $z$ to the fourth power gives modulus $9$ and argument $-5\pi/2$ . Division by $w$ then gives modulus $9/3=3$ and argument $-5\pi/2-\pi/12=-31\pi/12$ . Adding $2\pi$ normalises this to the principal argument $-7\pi/12$ .

Tier 3 · Hard

Mark scheme for CP-2.6 Tier 3 · Hard
Question	Scheme	Marks
1	$z=2e^{-5\pi i/7}$	4
(4 marks)		4
Notes
$(2e^{i\pi/7})^3=8e^{3\pi i/7}$ . Divide the right-hand side by this number: the modulus is $16/8=2$ and the argument is $-2\pi/7-3\pi/7=-5\pi/7$ . Thus $z=2e^{-5\pi i/7}$ .
2	Multiplying the equations gives $z_1^2=36e^{i\pi/3}$ . The two square roots are $6e^{i\pi/6}$ and $6e^{-5\pi i/6}$ . $\operatorname{Re}(z_1)>0$ selects $z_1=6e^{i\pi/6}$ . $\|z_2\|=12/6=2$ . $\arg z_2=5\pi/6-\pi/6=2\pi/3$ . Therefore $z_2=2e^{2\pi i/3}$ .	6
(6 marks)		6
Notes
Multiplying the two equations gives $z_1^2=36e^{i\pi/3}$ . Its two square roots are $6e^{i\pi/6}$ and $6e^{-5\pi i/6}$ . The real-part condition selects $z_1=6e^{i\pi/6}$ . Division into the product then gives modulus $12/6=2$ and argument $5\pi/6-\pi/6=2\pi/3$ , so $z_2=2e^{2\pi i/3}$ . The rejected branch would give negative real part and a second pair, so the stated condition is essential.
3	$\overline z=3e^{-\pi i/8}$ . $\|w\|=3^2/3=3$ . An argument of $w$ is $2(\pi/8)-(-\pi/8)=3\pi/8\in(-\pi,\pi]$ , so $w=3e^{3\pi i/8}$ with its principal argument. $\cos(3\pi/8)=\tfrac12\sqrt{2-\sqrt2}$ and $\sin(3\pi/8)=\tfrac12\sqrt{2+\sqrt2}$ . $w=\dfrac32\sqrt{2-\sqrt2}+\dfrac32\sqrt{2+\sqrt2}\,i$ .	5
(5 marks)		5
Notes
Conjugation reverses the argument. Squaring and division give modulus $3$ and argument $3\pi/8$ . Use the exact half-angle values at $3\pi/8$ , with both signs positive because its real and imaginary parts are positive, then multiply each component by $3$ .
4	Writing $\|z\|=r$ and $\|w\|=s$ , the modulus equations are $r^2s=32$ and $rs^2=16$ . Division gives $r/s=2$ , so substitution gives $s=2$ and $r=4$ . Write $\arg z=\alpha$ and $\arg w=\beta$ . The stated intervals make the principal equations $2\alpha+\beta=\pi/12$ and $\alpha+2\beta=-\pi/3$ . Solving these equations gives $\alpha=\pi/6$ and $\beta=-\pi/4$ . Therefore $z=4e^{i\pi/6}$ . Therefore $w=2e^{-i\pi/4}$ .	6
(6 marks)		6
Notes
Compare moduli to obtain two equations in the positive numbers $r$ and $s$ ; dividing them fixes their ratio before substitution. Compare arguments to obtain a different pair of linear equations. The supplied argument intervals keep both displayed argument sums within their principal representatives, so no extra multiple of $2\pi$ is possible.
5	Write $z=re^{i\theta}$ , so $\overline z=re^{-i\theta}$ . $z^3\overline z^2=r^5e^{i\theta}$ . Comparing moduli gives $r^5=32$ , so $r=2$ . Comparing principal arguments gives $\theta=-\pi/3$ . Thus $z=2\left[\cos\left(-\pi/3\right)+i\sin\left(-\pi/3\right)\right]$ . In Cartesian form, $z=2(1/2-i\sqrt3/2)=1-\sqrt3i$ .	6
(6 marks)		6
Notes
Conjugation reverses the argument but preserves the modulus. The mixed product therefore has modulus $r^5$ while its net argument is just $\theta$ . Compare these independently with the given modulus-argument form, then apply Euler's definition.

CP-2.7 · Construct and interpret simple loci in the Argand diagram such as |z - a| > r and arg(z - a) = theta.

Tier 1 · Easy

Mark scheme for CP-2.7 Tier 1 · Easy
Question	Scheme	Marks
1	The exterior of the circle with centre $(2,-1)$ and radius $3$ The circular boundary is not included	2
(2 marks)		2
Notes
$\|z-(2-i)\|$ is the distance from the point representing $z$ to $(2,-1)$ . Distances greater than $3$ lie outside the radius- $3$ circle, and the strict inequality excludes the circle itself.
2	Equal squared distances simplify to $2x-y=6$ . $\operatorname{Im}(z)=2$ gives $y=2$ and hence $x=4$ . Thus $z=4+2i$ , and both squared distances are $10$ .	3
(3 marks)		3
Notes
Squaring the equal distances gives $(x-1)^2+(y-1)^2=(x-5)^2+(y+1)^2$ , which simplifies to $2x-y=6$ . The second locus is $y=2$ , so $x=4$ . The point $(4,2)$ has squared distances $10$ from both $(1,1)$ and $(5,-1)$ , so it satisfies both loci. The two non-parallel lines have only this intersection.

Tier 2 · Standard

Mark scheme for CP-2.7 Tier 2 · Standard
Question	Scheme	Marks
1	$y=x+1$ with $x>-1$	3
(3 marks)		3
Notes
$z+1=(x+1)+iy$ is measured from the point $(-1,0)$ . An argument of $\pi/4$ gives $y=x+1$ , and both displacement components are positive, so $x>-1$ . The endpoint is excluded because its displacement is zero.
2	The two boundary circles have centre $(2,1)$ and radii $1$ and $3$ . The region is the annulus between these two circles. Both circular boundaries are included. Its area is $\pi(3^2-1^2)=8\pi$ .	4
(4 marks)		4
Notes
$\|z-(2+i)\|$ is the distance from $(2,1)$ . The compound inequality selects all points whose distance from $(2,1)$ is between $1$ and $3$ , inclusive, so the region is a closed annulus. Subtracting the area of the inner disc from the area of the outer disc gives $\pi(3^2)-\pi(1^2)=8\pi$ .
3	The distance comparison simplifies to the half-plane $2x+3y\geq-6$ , including its slanted boundary line. The modulus condition is $(x-2)^2+(y+1)^2<9$ , the interior of a circle whose boundary is excluded. At the origin, $0\geq-6$ and $(-2)^2+1^2=5<9$ , so the origin lies in the region.	3
(3 marks)		3
Notes
Square the two non-negative distances: $(x-2)^2+(y-1)^2\leq(x+2)^2+(y+5)^2$ reduces to $8x+12y\geq-24$ , or $2x+3y\geq-6$ . Translate the second modulus directly into the open disc centred at $(2,-1)$ . The strict and non-strict signs determine the two boundary conventions, and substitution tests the origin.

Tier 3 · Hard

Mark scheme for CP-2.7 Tier 3 · Hard
Question	Scheme	Marks
1	An open sector centred at $(1,-1)$ , of radius $4$ and angle $\pi/3$ Area $=\dfrac{8\pi}{3}$	5
(5 marks)		5
Notes
The modulus inequality gives the interior of the radius- $4$ circle centred at $(1,-1)$ . The argument inequalities select the sector between the rays of angles $0$ and $\pi/3$ , with both rays and the arc excluded. Its area is $\tfrac12r^2\theta=\tfrac12(4^2)(\pi/3)=8\pi/3$ ; excluding boundaries does not change the area.
2	The distance comparison simplifies to $x\geq5/2$ . The region is the closed segment of $\|z\|\leq5$ lying in $x\geq5/2$ . The chord endpoints are $(5/2,\pm5\sqrt3/2)$ . The central angle is $2\pi/3$ , giving sector area $25\pi/3$ . The enclosed isosceles triangle has area $25\sqrt3/4$ . The segment area is $25\pi/3-25\sqrt3/4$ .	6
(6 marks)		6
Notes
Writing $z=x+iy$ , the distance comparison gives $(x-5)^2+y^2\leq x^2+y^2$ , so $x\geq5/2$ . Both boundaries are included. Substituting $x=5/2$ into $x^2+y^2=25$ gives the chord endpoints $(5/2,\pm5\sqrt3/2)$ . The radii to these points subtend $2\cos^{-1}(1/2)=2\pi/3$ . The required segment is the sector of area $\tfrac12(5^2)(2\pi/3)=25\pi/3$ minus the isosceles triangle of area $\tfrac12(5^2)\sin(2\pi/3)=25\sqrt3/4$ . The point $(4,0)$ selects the side $x\geq5/2$ .
3	Writing $z=x+iy$ , the equal-distance locus simplifies to $y=x+3$ . The argument condition lies on the line $y=-x$ . The ray condition is $x<1$ . Solving the two line equations gives $x=-3/2$ . Then $y=3/2$ , so $z=-3/2+3i/2$ and $x<1$ confirms the required ray. The endpoint $z=1-i$ is excluded because $\arg(0)$ is undefined.	6
(6 marks)		6
Notes
Squaring the two distances gives $(x-2)^2+(y-1)^2=(x+2)^2+(y-5)^2$ , which reduces to $y=x+3$ . The argument is measured from $(1,-1)$ ; angle $3\pi/4$ gives the line $y+1=-(x-1)$ together with $x<1$ . Its intersection with the perpendicular bisector is $(-3/2,3/2)$ , and the restriction verifies the correct ray.
4	The centre is represented by $4+3i$ , so its distance from the origin is $5$ . A radius to a point of contact is perpendicular to its tangent, so each tangent length is $\sqrt{5^2-3^2}=4$ . The circle touches the real axis at the point represented by $4$ , giving one ray argument $0$ . If $\phi=\tan^{-1}(3/4)$ is the argument of $4+3i$ , symmetry about the line to the centre gives the other ray argument $2\phi$ . $\cos(2\phi)=7/25$ and $\sin(2\phi)=24/25$ . The other point of contact is represented by $4(7/25+24i/25)=28/25+96i/25$ , and its ray has argument $2\tan^{-1}(3/4)$ .	6
(6 marks)		6
Notes
Interpret the locus as a circle with centre $(4,3)$ and radius $3$ . A radius to a point of contact and its tangent form a right triangle with hypotenuse $5$ , so each tangent segment has length $4$ . One tangent is the real axis. The two tangents are symmetric about the line from the origin to the centre, so double that line's argument and use exact double-angle values to locate the second point.
5	The inequalities define closed discs with centres $(0,0)$ and $(1,1)$ and radii $2$ and $\sqrt2$ , respectively; the required region is their intersection. The boundary circles meet where $x^2+y^2=4$ and $(x-1)^2+(y-1)^2=2$ , which gives $x+y=2$ . The two intersection points are $(0,2)$ and $(2,0)$ . At the origin the radii to these points subtend angle $\pi/2$ , giving sector area $\pi$ . At $(1,1)$ the radii to the two points are opposite, giving a semicircle of area $\pi$ . The quadrilateral formed by the two centres and the two intersection points has area $2$ . Therefore the exact area of the intersection is $\pi+\pi-2=2\pi-2$ .	7
(7 marks)		7
Notes
Solve the two circle equations to locate their common chord. The lens is the sum of the quarter-disc sector at the origin and the semicircle sector at $(1,1)$ , less the two right triangles counted inside those sectors. Their combined area is the area- $2$ quadrilateral joining the centres to the chord endpoints.

CP-2.8 · Understand de Moivre's theorem and use it to find multiple angle formulae and sums of series.

Tier 1 · Easy

Mark scheme for CP-2.8 Tier 1 · Easy
Question	Scheme	Marks
1	$\cos5\theta+i\sin5\theta$	1
(1 mark)		1
Notes
Use de Moivre's theorem directly with $n=5$ : the modulus remains $1$ and the argument is multiplied by $5$ .
2	$1+i=\sqrt2(\cos(\pi/4)+i\sin(\pi/4))$ . By de Moivre's theorem, $(1+i)^8=16(\cos2\pi+i\sin2\pi)$ . Since $\cos2\pi=1$ and $\sin2\pi=0$ , $(1+i)^8=16$ .	3
(3 marks)		3
Notes
The modulus of $1+i$ is $\sqrt2$ and its principal argument is $\pi/4$ , so $1+i=\sqrt2(\cos(\pi/4)+i\sin(\pi/4))$ . De Moivre's theorem gives $(1+i)^8=(\sqrt2)^8(\cos(8\pi/4)+i\sin(8\pi/4))=16(\cos2\pi+i\sin2\pi)=16$ .

Tier 2 · Standard

Mark scheme for CP-2.8 Tier 2 · Standard
Question	Scheme	Marks
1	$\sin3\theta=3\sin\theta-4\sin^3\theta$	4
(4 marks)		4
Notes
Expand $(\cos\theta+i\sin\theta)^3$ . Equating imaginary parts with $\cos3\theta+i\sin3\theta$ gives $\sin3\theta=3\cos^2\theta\sin\theta-\sin^3\theta$ . Use $\cos^2\theta=1-\sin^2\theta$ to obtain $3\sin\theta-4\sin^3\theta$ .
2	$\sin3\theta=3\cos^2\theta\sin\theta-\sin^3\theta$ . $\cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta$ . Divide these identities to form $\tan3\theta$ . Dividing numerator and denominator by $\cos^3\theta$ gives $(3t-t^3)/(1-3t^2)$ . This holds where $\tan\theta$ and $\tan3\theta$ are defined.	5
(5 marks)		5
Notes
Equating parts in $(\cos\theta+i\sin\theta)^3$ gives $\sin3\theta=3\cos^2\theta\sin\theta-\sin^3\theta$ and $\cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta$ . Divide the first by the second and then numerator and denominator by $\cos^3\theta$ . With $t=\tan\theta$ , the result is $(3t-t^3)/(1-3t^2)$ . The derivation requires $\cos\theta\ne0$ and $\cos3\theta\ne0$ , matching the condition that the displayed tangents and quotient are defined.
3	$\displaystyle\int_0^\pi\sin^5\theta\,d\theta=\frac1{16}\left(\int_0^\pi\sin5\theta\,d\theta-5\int_0^\pi\sin3\theta\,d\theta+10\int_0^\pi\sin\theta\,d\theta\right)$ . $\displaystyle\int_0^\pi\sin5\theta\,d\theta=\frac25$ , $\displaystyle\int_0^\pi\sin3\theta\,d\theta=\frac23$ and $\displaystyle\int_0^\pi\sin\theta\,d\theta=2$ . The required value is $\displaystyle\frac1{16}\left(\frac25-\frac{10}{3}+20\right)$ . Therefore $\displaystyle\int_0^\pi\sin^5\theta\,d\theta=\frac{16}{15}$ .	4
(4 marks)		4
Notes
The supplied expansion follows by putting $z=e^{i\theta}$ , expanding $(z-z^{-1})^5$ , pairing opposite powers and using $z^m-z^{-m}=2i\sin(m\theta)$ . For the application, integrate the expansion term by term over $[0,\pi]$ . The three odd-frequency sine integrals are $2/5$ , $2/3$ and $2$ , giving $16/15$ after simplification.

Tier 3 · Hard

Mark scheme for CP-2.8 Tier 3 · Hard
Question	Scheme	Marks
1	$S=-\dfrac{\sqrt2}{2}$	5
(5 marks)		5
Notes
Let $q=\cos(\pi/4)+i\sin(\pi/4)=(1+i)/\sqrt2$ . Then $S$ is the real part of $1+q+\cdots+q^5=(1-q^6)/(1-q)$ . De Moivre gives $q^6=-i$ , so rationalising gives $\dfrac{1+i}{1-(1+i)/\sqrt2}$ , whose real part is $\dfrac{1-\sqrt2}{2-\sqrt2}=-\dfrac{\sqrt2}{2}$ . Hence $S=-\sqrt2/2$ .
2	Let $q=e^{i\theta}$ ; the required sum is the imaginary part of $q+q^2+\cdots+q^n$ . Since $q\ne1$ , $q+q^2+\cdots+q^n=\dfrac{q(1-q^n)}{1-q}$ . Use $1-e^{in\theta}=-2ie^{in\theta/2}\sin(n\theta/2)$ . Also $1-e^{i\theta}=-2ie^{i\theta/2}\sin(\theta/2)$ . The quotient is $\dfrac{\sin(n\theta/2)}{\sin(\theta/2)}e^{i(n+1)\theta/2}$ . Taking imaginary parts gives $\displaystyle\sum_{r=1}^{n}\sin r\theta=\dfrac{\sin(n\theta/2)\sin((n+1)\theta/2)}{\sin(\theta/2)}$ .	6
(6 marks)		6
Notes
Put $q=e^{i\theta}$ . The excluded values $\theta=2k\pi$ are exactly those for which $q=1$ and the geometric-series denominator vanishes. Otherwise, $\sum_{r=1}^{n}q^r=q(1-q^n)/(1-q)$ . Substitute $1-e^{i\alpha}=-2ie^{i\alpha/2}\sin(\alpha/2)$ into the numerator with $\alpha=n\theta$ and into the denominator with $\alpha=\theta$ . Cancelling the common factor gives $e^{i(n+1)\theta/2}\sin(n\theta/2)/\sin(\theta/2)$ . Its imaginary part is the required sum and is $\sin(n\theta/2)\sin((n+1)\theta/2)/\sin(\theta/2)$ .
3	Put $q=2e^{i\pi/6}=\sqrt3+i$ , so the required sum is the real part of $\displaystyle\sum_{r=1}^{7}q^r$ . $\displaystyle\sum_{r=1}^{7}q^r=\frac{q(1-q^7)}{1-q}$ . $q^7=-64\sqrt3-64i$ . Hence $q(1-q^7)=128+\sqrt3+(1+128\sqrt3)i$ and $1-q=1-\sqrt3-i$ . Rationalising gives $\displaystyle\sum_{r=1}^{7}q^r=-70-79\sqrt3+(10\sqrt3-39)i$ . Taking real parts gives $\displaystyle\sum_{r=1}^{7}2^r\cos\left(\frac{r\pi}{6}\right)=-70-79\sqrt3$ .	6
(6 marks)		6
Notes
Combine the cosine terms as the real part of the finite geometric series with ratio $q=2e^{i\pi/6}$ . Use $q(1-q^7)/(1-q)$ , then substitute $q=\sqrt3+i$ and $q^7=-64\sqrt3-64i$ . Rationalising the denominator produces $-70-79\sqrt3+(10\sqrt3-39)i$ , whose real part is the required sum.
4	Put $q=e^{i\theta}$ , so the required sum is the real part of $\sum_{r=1}^{n}rq^r$ . If $S=q+2q^2+\cdots+nq^n$ , then $qS=q^2+2q^3+\cdots+nq^{n+1}$ . Subtracting gives $(1-q)S=q+q^2+\cdots+q^n-nq^{n+1}$ . Using the finite geometric sum and dividing by $1-q$ gives $S=\dfrac{q-(n+1)q^{n+1}+nq^{n+2}}{(1-q)^2}$ . Divide numerator and denominator by $q$ , use $q+q^{-1}=2\cos\theta$ , and take real parts to obtain the stated identity. For $n=8$ and $\theta=2\pi/3$ , the numerator is $9\cos(16\pi/3)-8\cos6\pi-1=-27/2$ . The denominator is $2(1-\cos(2\pi/3))=3$ . Therefore $\displaystyle\sum_{r=1}^{8}r\cos(2r\pi/3)=-9/2$ .	8
(8 marks)		8
Notes
Write the arithmetico-geometric sum as $S=q+2q^2+\cdots+nq^n$ , multiply it by $q$ , and subtract. Replace the remaining geometric sum by its finite formula and solve for $S$ . Dividing numerator and denominator by $q$ makes the denominator real, so taking real parts produces the identity. The numerical substitution then reduces to exact cosine values.
5	For each listed angle, $4\theta=\pi/4+k\pi$ , so $\tan4\theta=1$ . Substitution in the supplied identity gives $4t(1-t^2)=1-6t^2+t^4$ . Rearranging gives $t^4+4t^3-6t^2-4t+1=0$ . The four listed tangent values are distinct. A quartic has at most four roots, so these are all the roots of the stated polynomial. By the product-of-roots form of Vieta's formulae, their product is $1$ .	6
(6 marks)		6
Notes
Each supplied angle makes $4\theta$ congruent to $\pi/4$ modulo $\pi$ . Insert $\tan4\theta=1$ into the supplied identity and rearrange to the stated quartic. The four distinct values exhaust its roots, and Vieta's formulae give their product directly from the constant and leading coefficients.

CP-2.9 · Know and use the definition e^(i theta) = cos theta + i sin theta and the form z = r e^(i theta).

Tier 1 · Easy

Mark scheme for CP-2.9 Tier 1 · Easy
Question	Scheme	Marks
1	$4e^{2\pi i/5}$	1
(1 mark)		1
Notes
Use $e^{i\theta}=\cos\theta+i\sin\theta$ with $\theta=2\pi/5$ , keeping the modulus $4$ .
2	$7e^{-i\pi/6}=\dfrac{7\sqrt3}{2}-\dfrac72i$ $\operatorname{Re}(z)=7\sqrt3/2$ and $\operatorname{Im}(z)=-7/2$	2
(2 marks)		2
Notes
Euler's definition gives $e^{-i\pi/6}=\cos(-\pi/6)+i\sin(-\pi/6)=\sqrt3/2-i/2$ . Multiplication by $7$ gives the stated Cartesian form, from which the two real components are read.

Tier 2 · Standard

Mark scheme for CP-2.9 Tier 2 · Standard
Question	Scheme	Marks
1	$e^{2i\theta}+e^{-2i\theta}=2\cos2\theta$	3
(3 marks)		3
Notes
Euler's definition gives $e^{2i\theta}=\cos2\theta+i\sin2\theta$ and $e^{-2i\theta}=\cos2\theta-i\sin2\theta$ . Adding cancels the imaginary parts and leaves $2\cos2\theta$ .
2	Put $u=e^{i\theta}$ to obtain $u^2-u+1=0$ . Thus $u=e^{i\pi/3}$ or $u=e^{-i\pi/3}$ . $e^{i\theta}=e^{i\pi/3}$ gives $\theta=\pi/3$ . $e^{i\theta}=e^{-i\pi/3}$ gives $\theta=5\pi/3$ in the stated interval.	4
(4 marks)		4
Notes
Put $u=e^{i\theta}$ . Then $u^2-u+1=0$ , so $u=(1\pm i\sqrt3)/2=e^{\pm i\pi/3}$ . In the specified interval, $e^{i\theta}=e^{i\pi/3}$ gives $\theta=\pi/3$ , while $e^{i\theta}=e^{-i\pi/3}$ gives $\theta=5\pi/3$ . Both values satisfy the original equation.
3	$2\cos5\theta\cos4\theta=\dfrac{(e^{5i\theta}+e^{-5i\theta})(e^{4i\theta}+e^{-4i\theta})}{2}$ . Expanding gives $\dfrac12(e^{9i\theta}+e^{i\theta}+e^{-i\theta}+e^{-9i\theta})$ . Regrouping gives $\dfrac{e^{9i\theta}+e^{-9i\theta}}2+\dfrac{e^{i\theta}+e^{-i\theta}}2$ . Applying the Euler identity to each pair gives $\cos9\theta+\cos\theta$ .	4
(4 marks)		4
Notes
Replace both cosine factors by their exponential pairs. Multiply the two binomials, group terms with opposite exponents, and convert each grouped pair back to a cosine.

Tier 3 · Hard

Mark scheme for CP-2.9 Tier 3 · Hard
Question	Scheme	Marks
1	$1+e^{i\theta}=2\cos(\theta/2)e^{i\theta/2}$	4
(4 marks)		4
Notes
Factor out $e^{i\theta/2}$ : $1+e^{i\theta}=e^{i\theta/2}(e^{-i\theta/2}+e^{i\theta/2})$ . By Euler's definition, the bracket is $2\cos(\theta/2)$ , giving the stated result. The interval ensures $\cos(\theta/2)>0$ , consistent with the displayed modulus.
2	Write $z=3e^{i\theta}$ , so $\overline z=3e^{-i\theta}$ . The equation becomes $e^{i(\theta+\pi/4)}=e^{-i\theta}$ . Hence $2\theta+\pi/4=2k\pi$ for some $k\in\mathbb Z$ . Thus $\theta=-\pi/8+k\pi$ , giving $-\pi/8$ and $7\pi/8$ in the principal interval. Therefore $z=3e^{-i\pi/8}$ or $z=3e^{7\pi i/8}$ .	5
(5 marks)		5
Notes
The modulus condition gives $z=3e^{i\theta}$ and therefore $\overline z=3e^{-i\theta}$ . Substitution and cancellation of the non-zero modulus give $e^{i(\theta+\pi/4)}=e^{-i\theta}$ , so the arguments differ by an integer multiple of $2\pi$ : $2\theta+\pi/4=2k\pi$ . Hence $\theta=-\pi/8+k\pi$ . Exactly two values lie in $(-\pi,\pi]$ , namely $-\pi/8$ and $7\pi/8$ , and both retain modulus $3$ .
3	Multiplying by $3+e^{-i\theta}$ gives the real denominator $10+6\cos\theta$ and adjusted numerator $6+10\cos\theta+8i\sin\theta$ . $\operatorname{Re}(w)=\dfrac{6+10\cos\theta}{10+6\cos\theta}$ and $\operatorname{Im}(w)=\dfrac{8\sin\theta}{10+6\cos\theta}$ . A purely imaginary value requires $\cos\theta=-3/5$ ; then the denominator is $32/5$ and $\sin\theta=\pm4/5$ . Let $\alpha=\tan^{-1}(4/3)$ , where $0<\alpha<\pi/2$ . When $\theta=\pi-\alpha$ , $\sin\theta=4/5$ and substitution gives $w=i$ . When $\theta=\pi+\alpha$ , $\sin\theta=-4/5$ and substitution gives $w=-i$ .	6
(6 marks)		6
Notes
Rationalise with the conjugate of the denominator. The squared modulus is $10+6\cos\theta$ , which is always positive, and expansion gives the displayed real and imaginary numerators. Setting the real numerator to zero produces a $3$ - $4$ - $5$ triangle. Treat the two angles separately so the sign of the sine is paired explicitly with the sign of the imaginary value.
4	$1-e^{i\theta}=e^{i\theta/2}(e^{-i\theta/2}-e^{i\theta/2})$ . Euler's definition gives $1-e^{i\theta}=-2ie^{i\theta/2}\sin(\theta/2)$ . Similarly, $1+e^{i\theta}=2e^{i\theta/2}\cos(\theta/2)$ . Division gives $\dfrac{1-e^{i\theta}}{1+e^{i\theta}}=-i\tan(\theta/2)$ . The equation therefore gives $\tan(\theta/2)=1/\sqrt3$ . Since $-\pi/2<\theta/2<\pi/2$ , the unique solution is $\theta=\pi/3$ .	6
(6 marks)		6
Notes
Factor the numerator and denominator symmetrically about the half-angle, then convert each exponential difference or sum with Euler's definition. The interval places the half-angle in the one-to-one branch of tangent.
5	Euler's definition gives $x=3\cos\theta$ and $y=\sin\theta$ . Therefore $x^2/9+y^2=\cos^2\theta+\sin^2\theta=1$ . On the real axis, $y=0$ , so $x^2/9=1$ . The real-axis intercepts are $(3,0)$ and $(-3,0)$ . On the imaginary axis, $x=0$ , so $y^2=1$ . The imaginary-axis intercepts are $(0,1)$ and $(0,-1)$ .	6
(6 marks)		6
Notes
Expand both exponentials into sine and cosine components and collect real and imaginary parts. Eliminate the parameter using $\cos^2\theta+\sin^2\theta=1$ . Set each coordinate equal to zero in turn and solve the resulting equation for the other coordinate.

CP-2.10 · Find the n distinct nth roots of r e^(i theta) for r != 0 and know that they form the vertices of a regular n-gon in the Argand diagram.

Tier 1 · Easy

Mark scheme for CP-2.10 Tier 1 · Easy
Question	Scheme	Marks
1	$2$ , $2e^{2\pi i/3}$ and $2e^{4\pi i/3}$	3
(3 marks)		3
Notes
Write $8=8e^{2m\pi i}$ . Each root has modulus $2$ and argument $2k\pi/3$ for $k=0,1,2$ , giving the three stated roots.
2	$\sqrt3+i=2e^{i\pi/6}$ . $w=(2e^{i\pi/6})^4=-8+8\sqrt3i$ .	2
(2 marks)		2
Notes
$\sqrt3+i=2e^{i\pi/6}$ . Raising this root to the fourth power gives $w=2^4e^{4i\pi/6}=16e^{2\pi i/3}=16(-1/2+i\sqrt3/2)=-8+8\sqrt3i$ .

Tier 2 · Standard

Mark scheme for CP-2.10 Tier 2 · Standard
Question	Scheme	Marks
1	$z=2e^{i\pi/12}$ , $2e^{7\pi i/12}$ , $2e^{13\pi i/12}$ or $2e^{19\pi i/12}$	4
(4 marks)		4
Notes
The root modulus is $16^{1/4}=2$ . The arguments are $(\pi/3+2k\pi)/4=\pi/12+k\pi/2$ for $k=0,1,2,3$ , which give the four listed values.
2	Both given vertices have modulus $2\sqrt2$ . Their arguments are $\pi/4$ and $\pi/2$ . The angle between adjacent vertices is $\pi/4=2\pi/n$ . Hence $n=8$ . The next vertex has argument $3\pi/4$ and is represented by $-2+2i$ .	5
(5 marks)		5
Notes
The given complex numbers are $2\sqrt2e^{i\pi/4}$ and $2\sqrt2e^{i\pi/2}$ . Their positive angular separation is $\pi/4$ . Since adjacent vertices of a regular $n$ -gon are separated by $2\pi/n$ , solve $2\pi/n=\pi/4$ to obtain $n=8$ . Continuing through another angle $\pi/4$ gives the vertex represented by $2\sqrt2e^{3\pi i/4}=-2+2i$ .
3	Every root has modulus $2$ . Their arguments before principal-value reduction are $\pi/18+k\pi/3$ for $k=0,1,\ldots,5$ . The only principal arguments in $(-\pi/3,\pi/3)$ are $\pi/18$ and $-5\pi/18$ . The selected roots are $2e^{i\pi/18}$ and $2e^{-5\pi i/18}$ . Their product is $4[\cos(-2\pi/9)+i\sin(-2\pi/9)]$ .	5
(5 marks)		5
Notes
Take sixth roots of the modulus and of the full argument family $\pi/3+2m\pi$ . Reduce the six resulting arguments to the principal interval, then apply the strict argument bound. Multiply the two selected moduli, add their arguments and give the product in the requested trigonometric modulus-argument form.

Tier 3 · Hard

Mark scheme for CP-2.10 Tier 3 · Hard
Question	Scheme	Marks
1	$z=2e^{i(-\pi/10+2k\pi/5)}$ for $k=0,1,2,3,4$ Area $=10\sin\left(\dfrac{2\pi}{5}\right)$	6
(6 marks)		6
Notes
Each root has modulus $32^{1/5}=2$ and argument $(-\pi/2+2k\pi)/5=-\pi/10+2k\pi/5$ . Joining each adjacent pair to the origin divides the pentagon into five congruent triangles, each with sides $2,2$ and included angle $2\pi/5$ . The total area is $5\times\tfrac12(2)(2)\sin(2\pi/5)=10\sin(2\pi/5)$ .
2	Adjacent eighth roots differ in argument by $\pi/4$ . Their positive real sum makes their arguments $-\pi/8$ and $\pi/8$ . $w=(2e^{-i\pi/8})^8=-256$ . Successive roots are obtained by rotations through $\pi/4$ . $z=2e^{i(-\pi/8+k\pi/4)}$ . The values $k=0,1,\ldots,7$ give all eight roots.	6
(6 marks)		6
Notes
Adjacent eighth roots differ in argument by $\pi/4$ . Because their sum is positive real, their arguments are symmetric about the positive real axis: $-\pi/8$ and $\pi/8$ . Their sum is $4\cos(\pi/8)=2\sqrt{2+\sqrt2}$ , which verifies the given value. Hence $w=(2e^{-i\pi/8})^8=256e^{-i\pi}=-256$ . Rotating successively through $\pi/4$ gives all roots $2e^{i(-\pi/8+k\pi/4)}$ for $k=0,\ldots,7$ .
3	The roots have modulus $2$ and arguments $\pi/16+k\pi/4$ . The four roots with positive real part have arguments $\pi/16$ , $5\pi/16$ , $-7\pi/16$ and $-3\pi/16$ . Pairing the first and fourth gives $4\cos(\pi/8)e^{-i\pi/16}$ . Pairing the second and third gives $4\cos(3\pi/8)e^{-i\pi/16}$ . Their sum is $4[\cos(\pi/8)+\cos(3\pi/8)]e^{-i\pi/16}$ . Therefore the exact sum is $4\sqrt2\cos(\pi/8)[\cos(-\pi/16)+i\sin(-\pi/16)]$ .	6
(6 marks)		6
Notes
The root arguments are $\pi/16+k\pi/4$ . Reduce them to the principal interval and select the four for which the cosine is positive. Pair angles symmetric about $-\pi/16$ : the half-differences are $\pi/8$ and $3\pi/8$ . The identity $e^{i(\alpha-\beta)}+e^{i(\alpha+\beta)}=2e^{i\alpha}\cos\beta$ , followed by $\cos(\pi/8)+\cos(3\pi/8)=\sqrt2\cos(\pi/8)$ , gives the result.
4	Every root has modulus $3$ and an argument $\pi/12+k\pi/3$ , where $k=0,1,\ldots,5$ . The greatest real part occurs for the root whose argument is $\pi/12$ . This root is $3e^{i\pi/12}$ . In Cartesian form it is $\dfrac34(\sqrt6+\sqrt2)+\dfrac34(\sqrt6-\sqrt2)i$ . Adjacent roots subtend angle $\pi/3$ at the origin, so their distance apart is $2(3)\sin(\pi/6)=3$ . The exact perimeter of the regular hexagon is therefore $6(3)=18$ .	6
(6 marks)		6
Notes
Take the sixth roots of $729e^{i(\pi/2+2m\pi)}$ and compare the cosines of their arguments to identify the greatest real part. Use the exact compound-angle values at $\pi/12$ for its Cartesian form. The six equally spaced roots form a regular hexagon whose side is the chord subtending $\pi/3$ .
5	The angle between adjacent roots is $2\pi/n$ . The side length is the chord $2(5)\sin(\pi/n)$ , so $\sin(\pi/n)=\sqrt{2-\sqrt2}/2=\sin(\pi/8)$ . Since $n\geq3$ , $0<\pi/n\leq\pi/3$ , so $n=8$ . $w=(5e^{i\pi/20})^8=390625e^{2\pi i/5}$ . Thus $w=\dfrac{390625}{4}[(\sqrt5-1)+i\sqrt{10+2\sqrt5}]$ . The four roots with positive imaginary part have arguments $\pi/20$ , $3\pi/10$ , $11\pi/20$ and $4\pi/5$ . The arguments add to $17\pi/10$ , whose principal equivalent is $-3\pi/10$ . Their product is $625\left[\cos\left(-3\pi/10\right)+i\sin\left(-3\pi/10\right)\right]$ .	8
(8 marks)		8
Notes
Relate the side to the chord subtended by one central angle; the domain for $n$ makes the exact sine value select $n=8$ . Raise the supplied root to the recovered power and use the exact values of cosine and sine at $2\pi/5$ . Select the four upper-half-plane roots, multiply their moduli and add their arguments before reducing to the principal interval.

CP-2.11 · Use complex roots of unity to solve geometric problems.

Tier 1 · Easy

Mark scheme for CP-2.11 Tier 1 · Easy
Question	Scheme	Marks
1	$\|1-\omega\|=\sqrt3$	2
(2 marks)		2
Notes
$\omega=-\tfrac12+\tfrac{\sqrt3}{2}i$ , so $1-\omega=\tfrac32-\tfrac{\sqrt3}{2}i$ . Its modulus is $\sqrt{(3/2)^2+(\sqrt3/2)^2}=\sqrt3$ .
2	The rotation multiplier is $\zeta=1/2+i\sqrt3/2$ . $\zeta(2-i)=1+\sqrt3/2+i(\sqrt3-1/2)$ .	2
(2 marks)		2
Notes
Rotation through $\pi/3$ is multiplication by a primitive sixth root of unity $\zeta=1/2+i\sqrt3/2$ . Thus $\zeta(2-i)=1+\sqrt3/2+i(\sqrt3-1/2)$ . Its modulus remains $\|2-i\|=\sqrt5$ , providing an independent rotation check.

Tier 2 · Standard

Mark scheme for CP-2.11 Tier 2 · Standard
Question	Scheme	Marks
1	$AB=BC=CA=\sqrt3$ , so triangle $ABC$ is equilateral	4
(4 marks)		4
Notes
From the easy chord calculation, $AB=\|1-\omega\|=\sqrt3$ . Multiplication by $\omega$ is a rotation through $2\pi/3$ and maps the displacement $1-\omega$ to $\omega-\omega^2$ , so $AB=BC$ . A second rotation gives $BC=CA$ . Hence all three sides are equal.
2	The coordinates are $A(1,0)$ , $B(1/2,\sqrt3/2)$ , $C(-1/2,\sqrt3/2)$ and $D(-1,0)$ . $BC\parallel AD$ because both are horizontal. $BC=1$ and $AD=2$ . $AB=CD=1$ , so the trapezium is isosceles. Its height is $\sqrt3/2$ . Its area is $\tfrac12(1+2)(\sqrt3/2)=3\sqrt3/4$ .	6
(6 marks)		6
Notes
The coordinates are $A(1,0)$ , $B(1/2,\sqrt3/2)$ , $C(-1/2,\sqrt3/2)$ and $D(-1,0)$ . Hence $BC$ and $AD$ are horizontal, with lengths $1$ and $2$ , so they are parallel. Also $AB=\|\zeta-1\|=1$ and $CD=\|\zeta^3-\zeta^2\|=\|\zeta^2(\zeta-1)\|=1$ , proving the trapezium is isosceles. Its perpendicular height is $\sqrt3/2$ , so its area is $\tfrac12(1+2)(\sqrt3/2)=3\sqrt3/4$ .
3	$z^6-1=(z-1)\displaystyle\prod_{k=1}^{5}(z-\zeta^k)$ . Dividing by $z-1$ gives $1+z+z^2+z^3+z^4+z^5=\displaystyle\prod_{k=1}^{5}(z-\zeta^k)$ . At $z=1$ , the left side of this identity is $1+1+1+1+1+1=6$ . The right side becomes $\displaystyle\prod_{k=1}^{5}(1-\zeta^k)$ , so the product of the distances is $\displaystyle\prod_{k=1}^{5}\left\|1-\zeta^k\right\|=\|6\|=6$ .	4
(4 marks)		4
Notes
Factor $z^6-1$ over its six roots of unity. Remove the factor $z-1$ corresponding to the trivial root; the quotient is $1+z+z^2+z^3+z^4+z^5$ . Evaluating the polynomial identity at $z=1$ gives $\prod_{k=1}^{5}(1-\zeta^k)=6$ ; each required distance is the modulus $\|1-\zeta^k\|$ , and the product of moduli is the modulus of the product.

Tier 3 · Hard

Mark scheme for CP-2.11 Tier 3 · Hard
Question	Scheme	Marks
1	Diagonal : side $=\dfrac{1+\sqrt5}{2}$	7
(7 marks)		7
Notes
Let $\zeta=e^{2\pi i/5}$ and put $u=\zeta+\zeta^{-1}=2\cos(2\pi/5)$ . Dividing $1+\zeta+\zeta^2+\zeta^3+\zeta^4=0$ by $\zeta^2$ gives $(\zeta^2+\zeta^{-2})+u+1=0$ . Since $\zeta^2+\zeta^{-2}=u^2-2$ , $u^2+u-1=0$ . As $u>0$ , $u=(\sqrt5-1)/2$ . A side has length $\|1-\zeta\|$ and a diagonal has length $\|1-\zeta^2\|=\|1-\zeta\|\|1+\zeta\|$ . Thus the ratio is $\|1+\zeta\|$ . Its square is $(1+\zeta)(1+\zeta^{-1})=2+u=(3+\sqrt5)/2$ , whose positive square root is $(1+\sqrt5)/2$ .
2	$\sum_{k=0}^{n-1}\zeta^k=0$ for $n\geq3$ . The conjugate sum $\sum_{k=0}^{n-1}\zeta^{-k}$ is also $0$ . $\|p-\zeta^k\|^2=\|p\|^2+1-p\zeta^{-k}-\overline p\zeta^k$ . Summing cancels the two root-of-unity terms. Therefore $\sum_{k=0}^{n-1}\|p-\zeta^k\|^2=n(\|p\|^2+1)$ . For $p=2+i$ , $\|p\|^2=5$ . For $n=6$ , the required sum is $6(5+1)=36$ .	7
(7 marks)		7
Notes
For every integer $n\geq3$ , $\zeta\ne1$ and the geometric sum gives $\sum_{k=0}^{n-1}\zeta^k=0$ ; its conjugate sum is also $0$ . For arbitrary $p\in\mathbb C$ , expand $\|p-\zeta^k\|^2=(p-\zeta^k)(\overline p-\zeta^{-k})=\|p\|^2+1-p\zeta^{-k}-\overline p\zeta^k$ . Summing cancels both root-of-unity sums, leaving $n(\|p\|^2+1)$ . With $p=2+i$ , $\|p\|^2=5$ , so the value is $6(6)=36$ .
3	The displacement from $P$ to $Q$ is $q-p=4+2i$ . Rotating this displacement through $\pm\pi/3$ about $P$ gives $r_1=p+\zeta(q-p)$ and $r_2=p+\overline\zeta(q-p)$ . $r_1=3-\sqrt3+i(2+2\sqrt3)$ . $r_2=3+\sqrt3+i(2-2\sqrt3)$ . The midpoint of $PQ$ is represented by $m=(p+q)/2=3+2i$ . $r_1-m=-\sqrt3+2\sqrt3i$ and $r_2-m=\sqrt3-2\sqrt3i=-(r_1-m)$ . Thus the half-turn about the midpoint $M$ maps $R_1$ to $R_2$ ; since $M$ lies on $PQ$ and neither equilateral vertex lies on $PQ$ , the two vertices lie on opposite sides of the line $PQ$ .	7
(7 marks)		7
Notes
Multiply $q-p$ by $\zeta$ and $\overline\zeta$ to rotate it through $\pm\pi/3$ about $P$ , then add $p$ . To prove the side statement without determinants, compare both new vertices with the midpoint of $PQ$ . Their midpoint displacements are negatives, so a half-turn about a point on the line swaps them. A non-degenerate equilateral vertex cannot lie on its base line, hence the swapped points occupy opposite sides.