Edexcel A-level Further Maths coverage

Complex numbers

Section CP-2
11 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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CP-2.1

Solve any quadratic equation with real coefficients. Solve cubic or quartic equations with real coefficients.

  • The roots of a real quadratic ax2+bx+c=0ax^2+bx+c=0 are x=b±b24ac2ax=\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 x3x^3 and xx 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

3 marks
ORIGINAL

Solve x24x+13=0x^2-4x+13=0, giving the roots in the form a+iba+ib.

Tier 2 · Standard

4 marks
ORIGINAL

Given that x=2x=2 is a root of x32x2+5x10=0x^3-2x^2+5x-10=0, solve the equation completely.

Tier 3 · Hard

5 marks
ORIGINAL

Given that x2+3x+6x^2+3x+6 is a factor of x4+3x2+36x^4+3x^2+36, solve x4+3x2+36=0x^4+3x^2+36=0 completely.

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+iyz=x+iy, the real and imaginary parts are Re(z)=x\operatorname{Re}(z)=x and Im(z)=y\operatorname{Im}(z)=y; the imaginary part is the real coefficient yy, not iyiy.
  • Add or subtract real parts together and imaginary parts together; when multiplying, expand and replace i2i^2 by 1-1.
  • 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 i2=1i^2=1 or to leave a complex number unsimplified instead of collecting it into the form x+iyx+iy.

Tier 1 · Easy

2 marks
ORIGINAL

Let z=(32i)+(5+7i)z=(3-2i)+(5+7i). Find zz, Re(z)\operatorname{Re}(z) and Im(z)\operatorname{Im}(z).

Tier 2 · Standard

3 marks
ORIGINAL

Express 4+i2i\dfrac{4+i}{2-i} in the form x+iyx+iy.

Tier 3 · Hard

5 marks
ORIGINAL

Express (2+3i)(14i)3+i\dfrac{(2+3i)(1-4i)}{3+i} in the form x+iyx+iy.

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+iyz=x+iy is z=xiy\overline z=x-iy, and zz=x2+y2=z2z\overline z=x^2+y^2=|z|^2 is real.
  • Changing ii to i-i throughout an expression with real coefficients shows that a non-real root a+iba+ib forces aiba-ib to be a root as well.
  • A conjugate pair produces the real quadratic factor [x(a+ib)][x(aib)]=(xa)2+b2[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

2 marks
ORIGINAL

For z=54iz=5-4i, write down z\overline z and calculate zzz\overline z.

Tier 2 · Standard

4 marks
ORIGINAL

The polynomial p(x)=x3x27x+15p(x)=x^3-x^2-7x+15 has real coefficients, and 2+i2+i is a root. Solve p(x)=0p(x)=0 completely.

Tier 3 · Hard

6 marks
ORIGINAL

A monic quartic polynomial has real coefficients. Two of its roots are 1+2i1+2i and 2+i-2+i. Determine the polynomial in expanded form.

CP-2.4

Use and interpret Argand diagrams.

  • On an Argand diagram, z=x+iyz=x+iy is represented by the point (x,y)(x,y), with the real axis horizontal and the imaginary axis vertical.
  • Differences of affixes represent displacement vectors, so z2z1|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+iyx+iy as (y,x)(y,x); label relevant points and exact coordinates on any sketch.

Tier 1 · Easy

2 marks
ORIGINAL

The complex number z=2+3iz=-2+3i is represented by PP on an Argand diagram. State the coordinates of PP and the quadrant in which it lies.

Tier 2 · Standard

3 marks
ORIGINAL

Points AA, BB and CC have affixes 1+i1+i, 5+i5+i and 1+4i1+4i. Determine the exact area of triangle ABCABC.

Tier 3 · Hard

5 marks
ORIGINAL

Points AA and BB have affixes 3+i3+i and 1+3i-1+3i, and OO is the origin. Prove that triangle OABOAB is right-angled and isosceles, and find its exact area.

CP-2.5

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

  • For z=x+iyz=x+iy, the modulus is r=z=x2+y2r=|z|=\sqrt{x^2+y^2} and an argument θ\theta satisfies tanθ=y/x\tan\theta=y/x with the quadrant checked.
  • The modulus-argument forms are r(cosθ+isinθ)r(\cos\theta+i\sin\theta) and, using Euler notation, reiθre^{i\theta}.
  • To return to Cartesian form, calculate x=rcosθx=r\cos\theta and y=rsinθy=r\sin\theta; exact special-angle values should remain exact.
  • Using tan1(y/x)\tan^{-1}(y/x) without a quadrant check can give an argument differing by π\pi; state the principal argument when requested.

Tier 1 · Easy

2 marks
ORIGINAL

Express 1+3i1+\sqrt3i in the form r(cosθ+isinθ)r(\cos\theta+i\sin\theta), where r>0r>0 and π<θπ-\pi<\theta\leq\pi.

Tier 2 · Standard

3 marks
ORIGINAL

Express 22i-2-2i in modulus-argument form using its principal argument.

Tier 3 · Hard

4 marks
ORIGINAL

The complex number z=a+ibz=a+ib has modulus 1010, positive real part and principal argument tan1(3/4)-\tan^{-1}(3/4). Determine zz in Cartesian form.

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 2π2\pi if a principal argument in (π,π](-\pi,\pi] is required.
  • Do not add moduli or multiply arguments; those operations do not correspond to complex multiplication.

Tier 1 · Easy

2 marks
ORIGINAL

Express 2eiπ/6×3eiπ/42e^{i\pi/6}\times3e^{i\pi/4} in the form reiθre^{i\theta}.

Tier 2 · Standard

3 marks
ORIGINAL

Express 12e5πi/63eπi/4\dfrac{12e^{5\pi i/6}}{3e^{-\pi i/4}} in the form reiθre^{i\theta} using the principal argument.

Tier 3 · Hard

4 marks
ORIGINAL

Given that z(2eiπ/7)3=16e2πi/7z(2e^{i\pi/7})^3=16e^{-2\pi i/7}, determine zz in the form reiθre^{i\theta}, where π<θπ-\pi<\theta\leq\pi.

CP-2.7

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

  • The condition za=r|z-a|=r is a circle of radius rr centred at the point with affix aa; << selects the interior and >> the exterior.
  • The condition arg(za)=θ\arg(z-a)=\theta is a ray starting at aa and making directed angle θ\theta with the positive real axis.
  • Translate z=x+iyz=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=az=a is excluded from an argument locus because the argument of zero is undefined.

Tier 1 · Easy

2 marks
ORIGINAL

Describe geometrically the locus z(2i)>3|z-(2-i)|>3, stating whether its boundary is included.

Tier 2 · Standard

3 marks
ORIGINAL

Let z=x+iyz=x+iy. Find a Cartesian description of the locus arg(z+1)=π/4\arg(z+1)=\pi/4, including the required restriction on xx.

Tier 3 · Hard

5 marks
ORIGINAL

A region is defined by z(1i)<4|z-(1-i)|<4 and 0<arg(z(1i))<π/30<\arg(z-(1-i))<\pi/3. Sketch the region and determine its exact 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 [cosθ+isinθ]n=cos(nθ)+isin(nθ)[\cos\theta+i\sin\theta]^n=\cos(n\theta)+i\sin(n\theta) for integer nn.
  • 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 eiθe^{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 mark
ORIGINAL

Use de Moivre's theorem to write (cosθ+isinθ)5(\cos\theta+i\sin\theta)^5 in modulus-argument form.

Tier 2 · Standard

4 marks
ORIGINAL

Starting from de Moivre's theorem, establish the identity sin3θ=3sinθ4sin3θ\sin3\theta=3\sin\theta-4\sin^3\theta.

Tier 3 · Hard

5 marks
ORIGINAL

Using a complex geometric series, evaluate exactly S=r=05cos(rπ4)S=\displaystyle\sum_{r=0}^{5}\cos\left(\dfrac{r\pi}{4}\right).

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 eiθ=cosθ+isinθe^{i\theta}=\cos\theta+i\sin\theta, so reiθre^{i\theta} has modulus rr and argument θ\theta.
  • The identities eiθ=cosθisinθe^{-i\theta}=\cos\theta-i\sin\theta, 2cosθ=eiθ+eiθ2\cos\theta=e^{i\theta}+e^{-i\theta} and 2isinθ=eiθeiθ2i\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 eiθe^{i\theta} as a real exponential: its modulus is 11, and its argument is periodic modulo 2π2\pi.

Tier 1 · Easy

1 mark
ORIGINAL

Express 4(cos2π5+isin2π5)4\left(\cos\dfrac{2\pi}{5}+i\sin\dfrac{2\pi}{5}\right) in exponential form.

Tier 2 · Standard

3 marks
ORIGINAL

Using Euler's definition, show that e2iθ+e2iθ=2cos2θe^{2i\theta}+e^{-2i\theta}=2\cos2\theta.

Tier 3 · Hard

4 marks
ORIGINAL

For π<θ<π-\pi<\theta<\pi, prove that 1+eiθ=2cos(θ/2)eiθ/21+e^{i\theta}=2\cos(\theta/2)e^{i\theta/2}.

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 zn=reiθz^n=re^{i\theta} with r>0r>0, the roots are z=r1/nei(θ+2kπ)/nz=r^{1/n}e^{i(\theta+2k\pi)/n} for k=0,1,,n1k=0,1,\ldots,n-1.
  • There are exactly nn distinct roots because their arguments differ by 2π/n2\pi/n before the pattern repeats.
  • On an Argand diagram the roots lie on the circle of radius r1/nr^{1/n} and form a regular nn-gon centred at the origin.
  • Using only the principal value of the argument gives just one root; include the 2kπ2k\pi term before dividing by nn.

Tier 1 · Easy

3 marks
ORIGINAL

Find the three cube roots of 88, giving them in exponential form.

Tier 2 · Standard

4 marks
ORIGINAL

Find the four roots of z4=16eiπ/3z^4=16e^{i\pi/3}, giving their arguments in the interval 0θ<2π0\leq\theta<2\pi.

Tier 3 · Hard

6 marks
ORIGINAL

The roots of z5=32eiπ/2z^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.

CP-2.11

Use complex roots of unity to solve geometric problems.

  • The nnth roots of unity are 1,ω,ω2,,ωn11,\omega,\omega^2,\ldots,\omega^{n-1}, where ω=e2πi/n\omega=e^{2\pi i/n}, and they are vertices of a regular nn-gon.
  • Because ωn=1\omega^n=1 and ω1\omega\ne1, the geometric sum gives 1+ω++ωn1=01+\omega+\cdots+\omega^{n-1}=0.
  • Distances are moduli of differences, while multiplying every affix by ω\omega rotates the entire figure through 2π/n2\pi/n without changing lengths.
  • Do not cancel ω1\omega-1 in ωn1=(ω1)(1++ωn1)\omega^n-1=(\omega-1)(1+\cdots+\omega^{n-1}) without first stating that the chosen root is not 11.

Tier 1 · Easy

2 marks
ORIGINAL

Let ω=e2πi/3\omega=e^{2\pi i/3}. Find the exact distance between the points with affixes 11 and ω\omega.

Tier 2 · Standard

4 marks
ORIGINAL

Let ω=e2πi/3\omega=e^{2\pi i/3}. Points AA, BB and CC have affixes 11, ω\omega and ω2\omega^2. Prove that triangle ABCABC is equilateral.

Tier 3 · Hard

7 marks
ORIGINAL

The fifth roots of unity are the vertices of a regular pentagon. Prove that the ratio of a diagonal to a side is 1+52\dfrac{1+\sqrt5}{2}.