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LearnYFM01WFM02 · Further Pure Mathematics 2Further complex numbersPolar and exponential forms
Further complex numbers · Subtopic 1 of 2

Polar and Exponential Forms

Write zz as r(cos⁡θ+isin⁡θ)r(\cos\theta+i\sin\theta) or reiθre^{i\theta}. In these forms multiplying is easy: multiply the moduli and add the arguments. Dividing: divide the moduli and subtract the arguments.

Cambridge 9709 Paper 3Edexcel YFM01 WFM02Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 40 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Convert between x+iyx+iy, r(cos⁡θ+isin⁡θ)r(\cos\theta+i\sin\theta) and reiθre^{i\theta}.
  • Multiply and divide in polar form, and interpret the result as a rotation and enlargement.
  • Find powers of a complex number using polar form, and reduce the argument to the principal range.
Pause and think

In the first subtopic, multiplying by ii rotated a point by 90∘90^\circ and kept its distance from OO. What do you think multiplying by 2i2i does?

Section 2 of 14

Core idea

Start hereEuler's jewel

eiπ+1=0e^{i\pi}+1=0 links the five most important numbers in mathematics. It comes from Euler's formula eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta, which turns rotations into exponentials, so that complex multiplication obeys ordinary index laws.

The big idea

eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta, so z=reiθz=re^{i\theta}. Then r1eiα⋅r2eiβ=r1r2ei(α+β)r_1e^{i\alpha}\cdot r_2e^{i\beta}=r_1r_2e^{i(\alpha+\beta)} and (reiθ)n=rneinθ(re^{i\theta})^n=r^ne^{in\theta}.

1. Why Euler's formula is plausible

Same behaviour as an exponential
  1. f(θ)=cos⁡θ+isin⁡θf(\theta)=\cos\theta+i\sin\theta
    Define.
  2. f′(θ)=−sin⁡θ+icos⁡θ=i(cos⁡θ+isin⁡θ)=if(θ)f'(\theta)=-\sin\theta+i\cos\theta=i(\cos\theta+i\sin\theta)=if(\theta)
    Differentiate; factor ii (using i2=−1i^2=-1).
  3. f′=if, f(0)=1 ⇒ f(θ)=eiθf'=if,\ f(0)=1\ \Rightarrow\ f(\theta)=e^{i\theta}
    The function with derivative kk times itself and value 11 at 00 is ekθe^{k\theta}.
z=x+iy=r(cos⁡θ+isin⁡θ)=reiθz=x+iy=r(\cos\theta+i\sin\theta)=re^{i\theta}

2. Powers

Powers of 1 + i spiral outwards
−6−4−224−4−3−2−1120ReIm45°1 + i(1 + i)² = 2i(1 + i)³(1 + i)⁴ = −4(1 + i)⁵
1+i=2 eiπ/41+i=\sqrt2\,e^{i\pi/4}. Each multiplication by it turns the arrow another 45∘45^\circ and stretches it by 2\sqrt2.
Worked example(1+i)8(1+i)^8

Evaluate.

Goal
Simplest form.
Strategy
Exponential form, then index laws.
  1. 1+i=2eiπ/41+i=\sqrt2e^{i\pi/4}
    r=2r=\sqrt2, θ=π4\theta=\frac\pi4.
  2. (1+i)8=(2)8ei⋅2π=16⋅1(1+i)^8=(\sqrt2)^8e^{i\cdot2\pi}=16\cdot1
    e2πi=cos⁡2π+isin⁡2π=1e^{2\pi i}=\cos2\pi+i\sin2\pi=1.
Answer1616.
✓ Check(1+i)2=2i(1+i)^2=2i, so (1+i)8=(2i)4=16i4=16(1+i)^8=(2i)^4=16i^4=16 ✓.
Worked example4eiπ/32e−iπ/6\dfrac{4e^{i\pi/3}}{2e^{-i\pi/6}}

Write as x+iyx+iy.

Goal
Cartesian form.
Strategy
Divide moduli; subtract args.
  1. 2ei(π/3+π/6)=2eiπ/22e^{i(\pi/3+\pi/6)}=2e^{i\pi/2}
    Index laws.
  2. =2(cos⁡π2+isin⁡π2)=2i=2(\cos\tfrac\pi2+i\sin\tfrac\pi2)=2i
    Euler.
Answer2i2i.
✓ CheckModulus 22, argument π2\frac\pi2: on the positive imaginary axis ✓.
Pause and think

Write −2i-2i as reiθre^{i\theta}.

Summary

  • eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta.
  • Index laws for products and powers.
Key words
Exponential form
reiθre^{i\theta}.
HistoryThe most beautiful equation. Leonhard Euler published eix=cos⁡x+isin⁡xe^{ix}=\cos x+i\sin x in 1748. Putting x=πx=\pi gives eiπ+1=0e^{i\pi}+1=0, which links five of the most important numbers in mathematics (ee, ii, π\pi, 11 and 00) in one line. Surveys of mathematicians regularly vote it the most beautiful equation ever written.
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