CP-2 Complex numbers — printable coverage pack

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

The roots of a real quadratic $ax^2+bx+c=0$ are $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ ; a negative discriminant produces a complex-conjugate pair.
For a cubic, first look for a real linear factor using inspection, the factor theorem or a supplied root, then solve the remaining quadratic.
A useful quartic principle is to recognise a biquadratic or factor the expression into two real quadratics; for example, symmetric $x^3$ and $x$ coefficients can guide the signs of the linear terms.
Do not stop after finding one factor: solve every remaining quadratic and list every root, including non-real roots.

Tier 1 · Easy

1. Solve $x^2-4x+13=0$ , giving the roots in the form $a+ib$ .[3 marks]
Answer
- $x=2+3i$ or $x=2-3i$
Method: Complete the square: $x^2-4x+13=(x-2)^2+9$ . Hence $(x-2)^2=-9$ , so $x-2=\pm3i$ and therefore $x=2\pm3i$ .

Tier 2 · Standard

1. Given that $x=2$ is a root of $x^3-2x^2+5x-10=0$ , solve the equation completely.[4 marks]
Answer
- $x=2$ , $x=i\sqrt5$ or $x=-i\sqrt5$
Method: 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 $x=\pm i\sqrt5$ .

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 marks]
Answer
- $x=\dfrac{-3+i\sqrt{15}}2$ , $x=\dfrac{-3-i\sqrt{15}}2$ , $x=\dfrac{3+i\sqrt{15}}2$ or $x=\dfrac{3-i\sqrt{15}}2$
Method: 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 $x=(3\pm i\sqrt{15})/2$ .

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'.

For $z=x+iy$ , the real and imaginary parts are $\operatorname{Re}(z)=x$ and $\operatorname{Im}(z)=y$ ; the imaginary part is the real coefficient $y$ , not $iy$ .
Add or subtract real parts together and imaginary parts together; when multiplying, expand and replace $i^2$ by $-1$ .
To divide Cartesian complex numbers, multiply numerator and denominator by the conjugate of the denominator so that the denominator becomes real.
A common error is to use $i^2=1$ or to leave a complex number unsimplified instead of collecting it into the form $x+iy$ .

Tier 1 · Easy

1. Let $z=(3-2i)+(5+7i)$ . Find $z$ , $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ .[2 marks]
Answer
- $z=8+5i$
- $\operatorname{Re}(z)=8$
- $\operatorname{Im}(z)=5$
Method: Add corresponding parts: $z=(3+5)+(-2+7)i=8+5i$ . Therefore the real part is $8$ and the imaginary part is $5$ .

Tier 2 · Standard

1. Express $\dfrac{4+i}{2-i}$ in the form $x+iy$ .[3 marks]
Answer
- $\dfrac75+\dfrac65i$
Method: Multiply by the conjugate $2+i$ : $\dfrac{4+i}{2-i}=\dfrac{(4+i)(2+i)}{(2-i)(2+i)}=\dfrac{7+6i}{5}=\dfrac75+\dfrac65i$ .

Tier 3 · Hard

1. Express $\dfrac{(2+3i)(1-4i)}{3+i}$ in the form $x+iy$ .[5 marks]
Answer
- $\dfrac{37}{10}-\dfrac{29}{10}i$
Method: First expand the numerator: $(2+3i)(1-4i)=2-8i+3i-12i^2=14-5i$ . Then multiply by the conjugate of the denominator: $\dfrac{14-5i}{3+i}=\dfrac{(14-5i)(3-i)}{10}=\dfrac{37-29i}{10}$ .

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

The conjugate of $z=x+iy$ is $\overline z=x-iy$ , and $z\overline z=x^2+y^2=|z|^2$ is real.
Changing $i$ to $-i$ throughout an expression with real coefficients shows that a non-real root $a+ib$ forces $a-ib$ to be a root as well.
A conjugate pair produces the real quadratic factor $[x-(a+ib)][x-(a-ib)]=(x-a)^2+b^2$ .
The conjugate-root rule requires real polynomial coefficients; applying it without checking this condition is a common error.

Tier 1 · Easy

1. For $z=5-4i$ , write down $\overline z$ and calculate $z\overline z$ .[2 marks]
Answer
- $\overline z=5+4i$
- $z\overline z=41$
Method: Reverse the sign of the imaginary part to get $\overline z=5+4i$ . Then $z\overline z=(5-4i)(5+4i)=5^2+4^2=41$ .

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 marks]
Answer
- $x=2+i$ , $x=2-i$ or $x=-3$
Method: 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 $-3$ .

Tier 3 · Hard

1. A monic quartic polynomial has real coefficients. Two of its roots are $1+2i$ and $-2+i$ . Determine the polynomial in expanded form.[6 marks]
Answer
- $x^4+2x^3+2x^2+10x+25$
Method: The other roots are $1-2i$ and $-2-i$ . The first conjugate pair gives $(x-1)^2+4=x^2-2x+5$ , and the second gives $(x+2)^2+1=x^2+4x+5$ . Multiplying yields $(x^2-2x+5)(x^2+4x+5)=x^4+2x^3+2x^2+10x+25$ .

CP-2.4 · Use and interpret Argand diagrams.

On an Argand diagram, $z=x+iy$ is represented by the point $(x,y)$ , with the real axis horizontal and the imaginary axis vertical.
Differences of affixes represent displacement vectors, so $|z_2-z_1|$ is the distance between the corresponding points.
Coordinate geometry applies directly: use gradients and scalar products to establish angles, and split polygons into triangles with base-and-height reasoning to establish areas.
Do not interchange the axes or plot $x+iy$ as $(y,x)$ ; label relevant points and exact coordinates on any sketch.

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 marks]
Answer
- $P=(-2,3)$
- Second quadrant
Method: 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.

Tier 2 · Standard

1. Points $A$ , $B$ and $C$ have affixes $1+i$ , $5+i$ and $1+4i$ . Determine the exact area of triangle $ABC$ .[3 marks]
Answer
- $6$ square units
Method: 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 $\tfrac12\times4\times3=6$ .

Tier 3 · Hard

1. Points $A$ and $B$ have affixes $3+i$ and $-1+3i$ , and $O$ is the origin. Prove that triangle $OAB$ is right-angled and isosceles, and find its exact area.[5 marks]
Answer
- $OA=OB=\sqrt{10}$
- $OA\perp OB$
- Area $=5$ square units
Method: 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 $\tfrac12(\sqrt{10})(\sqrt{10})=5$ .

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

For $z=x+iy$ , the modulus is $r=|z|=\sqrt{x^2+y^2}$ and an argument $\theta$ satisfies $\tan\theta=y/x$ with the quadrant checked.
The modulus-argument forms are $r(\cos\theta+i\sin\theta)$ and, using Euler notation, $re^{i\theta}$ .
To return to Cartesian form, calculate $x=r\cos\theta$ and $y=r\sin\theta$ ; exact special-angle values should remain exact.
Using $\tan^{-1}(y/x)$ without a quadrant check can give an argument differing by $\pi$ ; state the principal argument when requested.

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 marks]
Answer
- $2\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)$
Method: The modulus is $\sqrt{1^2+(\sqrt3)^2}=2$ . The point is in the first quadrant and $\tan\theta=\sqrt3$ , so $\theta=\pi/3$ .

Tier 2 · Standard

1. Express $-2-2i$ in modulus-argument form using its principal argument.[3 marks]
Answer
- $2\sqrt2\left(\cos\left(-\dfrac{3\pi}{4}\right)+i\sin\left(-\dfrac{3\pi}{4}\right)\right)$
Method: 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 $r(\cos\theta+i\sin\theta)$ .

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 marks]
Answer
- $z=8-6i$
Method: 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 $z=10(4/5-3i/5)=8-6i$ .

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

When multiplying complex numbers in modulus-argument form, multiply their moduli and add their arguments.
When dividing, divide the moduli and subtract the denominator's argument from the numerator's argument.
After calculating, add or subtract $2\pi$ if a principal argument in $(-\pi,\pi]$ is required.
Do not add moduli or multiply arguments; those operations do not correspond to complex multiplication.

Tier 1 · Easy

1. Express $2e^{i\pi/6}\times3e^{i\pi/4}$ in the form $re^{i\theta}$ .[2 marks]
Answer
- $6e^{5\pi i/12}$
Method: 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 $6e^{5\pi i/12}$ .

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 marks]
Answer
- $4e^{-11\pi i/12}$
Method: 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 $-11\pi/12$ .

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 marks]
Answer
- $z=2e^{-5\pi i/7}$
Method: $(2e^{i\pi/7})^3=8e^{3\pi i/7}$ . 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 $z=2e^{-5\pi i/7}$ .

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

The condition $|z-a|=r$ is a circle of radius $r$ centred at the point with affix $a$ ; $<$ selects the interior and $>$ the exterior.
The condition $\arg(z-a)=\theta$ is a ray starting at $a$ and making directed angle $\theta$ with the positive real axis.
Translate $z=x+iy$ when an equation is needed: modulus conditions become distance equations and argument conditions give a line plus a direction restriction.
Strict inequalities exclude their boundary, and the point $z=a$ is excluded from an argument locus because the argument of zero is undefined.

Tier 1 · Easy

1. Describe geometrically the locus $|z-(2-i)|>3$ , stating whether its boundary is included.[2 marks]
Answer
- The exterior of the circle with centre $(2,-1)$ and radius $3$
- The circular boundary is not included
Method: $|z-(2-i)|$ 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.

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 marks]
Answer
- $y=x+1$ with $x>-1$
Method: $z+1=(x+1)+iy$ is measured from the point $(-1,0)$ . An argument of $\pi/4$ gives $y=x+1$ , but it is the ray in the first-quadrant direction from $(-1,0)$ , so $x>-1$ . The endpoint is excluded because its displacement is zero.

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 marks]
Answer
- An open sector centred at $(1,-1)$ , of radius $4$ and angle $\pi/3$
- Area $=\dfrac{8\pi}{3}$
Method: 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.

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

De Moivre's theorem states $[\cos\theta+i\sin\theta]^n=\cos(n\theta)+i\sin(n\theta)$ for integer $n$ .
To derive a multiple-angle identity, expand the left-hand side and equate real parts for cosine or imaginary parts for sine.
For trigonometric sums, write the terms as real or imaginary parts of a geometric series in $e^{i\theta}$ , sum it, then take the required part.
A common error is to equate the whole binomial expansion with only a real or only an imaginary target; separate the two parts first.

Tier 1 · Easy

1. Use de Moivre's theorem to write $(\cos\theta+i\sin\theta)^5$ in modulus-argument form.[1 mark]
Answer
- $\cos5\theta+i\sin5\theta$
Method: Apply de Moivre's theorem directly with $n=5$ : the modulus remains $1$ and the argument is multiplied by $5$ .

Tier 2 · Standard

1. Starting from de Moivre's theorem, establish the identity $\sin3\theta=3\sin\theta-4\sin^3\theta$ .[4 marks]
Answer
- $\sin3\theta=3\sin\theta-4\sin^3\theta$
Method: 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 $3\sin\theta-4\sin^3\theta$ .

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 marks]
Answer
- $S=-\dfrac{\sqrt2}{2}$
Method: 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 $S=-\sqrt2/2$ .

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

Euler's definition is $e^{i\theta}=\cos\theta+i\sin\theta$ , so $re^{i\theta}$ has modulus $r$ and argument $\theta$ .
The identities $e^{-i\theta}=\cos\theta-i\sin\theta$ , $2\cos\theta=e^{i\theta}+e^{-i\theta}$ and $2i\sin\theta=e^{i\theta}-e^{-i\theta}$ follow immediately.
Exponential form makes rotations and trigonometric rearrangements concise; for example, factor out the exponential with the mean of two arguments.
Do not interpret $e^{i\theta}$ as a real exponential: its modulus is $1$ , and its argument is periodic modulo $2\pi$ .

Tier 1 · Easy

1. Express $4\left(\cos\dfrac{2\pi}{5}+i\sin\dfrac{2\pi}{5}\right)$ in exponential form.[1 mark]
Answer
- $4e^{2\pi i/5}$
Method: Use $e^{i\theta}=\cos\theta+i\sin\theta$ with $\theta=2\pi/5$ , keeping the modulus $4$ .

Tier 2 · Standard

1. Using Euler's definition, show that $e^{2i\theta}+e^{-2i\theta}=2\cos2\theta$ .[3 marks]
Answer
- $e^{2i\theta}+e^{-2i\theta}=2\cos2\theta$
Method: 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 $2\cos2\theta$ .

Tier 3 · Hard

1. For $-\pi<\theta<\pi$ , prove that $1+e^{i\theta}=2\cos(\theta/2)e^{i\theta/2}$ .[4 marks]
Answer
- $1+e^{i\theta}=2\cos(\theta/2)e^{i\theta/2}$
Method: 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.

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.

If $z^n=re^{i\theta}$ with $r>0$ , the roots are $z=r^{1/n}e^{i(\theta+2k\pi)/n}$ for $k=0,1,\ldots,n-1$ .
There are exactly $n$ distinct roots because their arguments differ by $2\pi/n$ before the pattern repeats.
On an Argand diagram the roots lie on the circle of radius $r^{1/n}$ and form a regular $n$ -gon centred at the origin.
Using only the principal value of the argument gives just one root; include the $2k\pi$ term before dividing by $n$ .

Tier 1 · Easy

1. Find the three cube roots of $8$ , giving them in exponential form.[3 marks]
Answer
- $2$ , $2e^{2\pi i/3}$ and $2e^{4\pi i/3}$
Method: 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.

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 marks]
Answer
- $z=2e^{i\pi/12}$ , $2e^{7\pi i/12}$ , $2e^{13\pi i/12}$ or $2e^{19\pi i/12}$
Method: 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.

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 marks]
Answer
- $z=2e^{i(-\pi/10+2k\pi/5)}$ for $k=0,1,2,3,4$
- Area $=10\sin\left(\dfrac{2\pi}{5}\right)$
Method: 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 $5\times\tfrac12(2)(2)\sin(2\pi/5)=10\sin(2\pi/5)$ .

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

The $n$ th roots of unity are $1,\omega,\omega^2,\ldots,\omega^{n-1}$ , where $\omega=e^{2\pi i/n}$ , and they are vertices of a regular $n$ -gon.
Because $\omega^n=1$ and $\omega\ne1$ , the geometric sum gives $1+\omega+\cdots+\omega^{n-1}=0$ .
Distances are moduli of differences, while multiplying every affix by $\omega$ rotates the entire figure through $2\pi/n$ without changing lengths.
Do not cancel $\omega-1$ in $\omega^n-1=(\omega-1)(1+\cdots+\omega^{n-1})$ without first stating that the chosen root is not $1$ .

Tier 1 · Easy

1. Let $\omega=e^{2\pi i/3}$ . Find the exact distance between the points with affixes $1$ and $\omega$ .[2 marks]
Answer
- $|1-\omega|=\sqrt3$
Method: $\omega=-\tfrac12+\tfrac{\sqrt3}{2}i$ , so $1-\omega=\tfrac32-\tfrac{\sqrt3}{2}i$ . Its modulus is $\sqrt{(3/2)^2+(\sqrt3/2)^2}=\sqrt3$ .

Tier 2 · Standard

1. Let $\omega=e^{2\pi i/3}$ . Points $A$ , $B$ and $C$ have affixes $1$ , $\omega$ and $\omega^2$ . Prove that triangle $ABC$ is equilateral.[4 marks]
Answer
- $AB=BC=CA=\sqrt3$ , so triangle $ABC$ is equilateral
Method: 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.

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 marks]
Answer
- Diagonal : side $=\dfrac{1+\sqrt5}{2}$
Method: 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 $(1+\sqrt5)/2$ .