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Exponential form of a complex numberAQA A-Level Further Maths: Revision notes

Section 1

The definition eiθ=cos⁡θ+isin⁡θ\mathrm{e}^{\mathrm{i}\theta}=\cos\theta+\mathrm{i}\sin\theta

For real θ\theta (in radians), the exponential form is defined by eiθ=cos⁡θ+isin⁡θ.\mathrm{e}^{\mathrm{i}\theta}=\cos\theta+\mathrm{i}\sin\theta. This is a complex number with modulus 11 and argument θ\theta. Every complex number z≠0z\ne0 can then be written z=reiθ,r=∣z∣, θ=arg⁡z.z=r\mathrm{e}^{\mathrm{i}\theta},\quad r=|z|,\ \theta=\arg z. This is the modulus-argument form r(cos⁡θ+isin⁡θ)r(\cos\theta+\mathrm{i}\sin\theta) written compactly. Because cos⁡\cos and sin⁡\sin repeat every 2π2\pi, ei(θ+2π)=eiθ\mathrm{e}^{\mathrm{i}(\theta+2\pi)}=\mathrm{e}^{\mathrm{i}\theta}. Special values: ei0=1\mathrm{e}^{\mathrm{i}0}=1, eiπ/2=i\mathrm{e}^{\mathrm{i}\pi/2}=\mathrm{i}, eiπ=−1\mathrm{e}^{\mathrm{i}\pi}=-1, e3iπ/2=−i\mathrm{e}^{3\mathrm{i}\pi/2}=-\mathrm{i}.

Key termsexponential formmodulusargument
Exam tip

eiπ+1=0\mathrm{e}^{\mathrm{i}\pi}+1=0: from the definition, cos⁡π+isin⁡π=−1\cos\pi+\mathrm{i}\sin\pi=-1.

Section 2

Converting between forms

To Cartesian: reiθ=rcos⁡θ+i rsin⁡θr\mathrm{e}^{\mathrm{i}\theta}=r\cos\theta+\mathrm{i}\,r\sin\theta. For example 2eiπ/3=2(12+32i)=1+3 i2\mathrm{e}^{\mathrm{i}\pi/3}=2\left(\frac12+\frac{\sqrt3}{2}\mathrm{i}\right)=1+\sqrt3\,\mathrm{i}. To exponential: find r=a2+b2r=\sqrt{a^2+b^2}, then θ\theta using the quadrant. For 1−3 i1-\sqrt3\,\mathrm{i}: r=2r=2 and the point is in the fourth quadrant with tan⁡θ=−3\tan\theta=-\sqrt3, so θ=−π3\theta=-\frac{\pi}{3} and z=2e−iπ/3z=2\mathrm{e}^{-\mathrm{i}\pi/3}. The conjugate of reiθr\mathrm{e}^{\mathrm{i}\theta} is re−iθr\mathrm{e}^{-\mathrm{i}\theta}, since cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta and sin⁡(−θ)=−sin⁡θ\sin(-\theta)=-\sin\theta.

Key termsconjugate
Common mistake

Using tan⁡−1ba\tan^{-1}\frac{b}{a} without checking the quadrant. For −1+i-1+\mathrm{i} the argument is 3π4\frac{3\pi}{4}, not −π4-\frac{\pi}{4}.

Section 3

Multiplying, dividing and powers

The exponential form follows the laws of indices. For z1=r1eiθ1z_1=r_1\mathrm{e}^{\mathrm{i}\theta_1} and z2=r2eiθ2z_2=r_2\mathrm{e}^{\mathrm{i}\theta_2}: z1z2=r1r2ei(θ1+θ2),z1z2=r1r2ei(θ1−θ2),z1n=r1neinθ1.z_1z_2=r_1r_2\mathrm{e}^{\mathrm{i}(\theta_1+\theta_2)},\quad \frac{z_1}{z_2}=\frac{r_1}{r_2}\mathrm{e}^{\mathrm{i}(\theta_1-\theta_2)},\quad z_1^n=r_1^n\mathrm{e}^{\mathrm{i}n\theta_1}. Multiply the moduli, add the arguments; divide the moduli, subtract the arguments. Example: w1=3eiπ/4w_1=3\mathrm{e}^{\mathrm{i}\pi/4}, w2=2e−iπ/4w_2=2\mathrm{e}^{-\mathrm{i}\pi/4}. Then w1w2=6e0=6w_1w_2=6\mathrm{e}^{0}=6 and w1w2=32eiπ/2=32i\frac{w_1}{w_2}=\frac32\mathrm{e}^{\mathrm{i}\pi/2}=\frac32\mathrm{i}.

Key termsmultiply moduli, add arguments
Common mistake

Adding the moduli in a product, or forgetting to raise the modulus to the power nn as well as multiplying the argument by nn.

Section 4

Real and imaginary results

A complex number reiθr\mathrm{e}^{\mathrm{i}\theta} is real when θ\theta is a multiple of π\pi (positive real if θ=2kπ\theta=2k\pi, negative real if θ=(2k+1)π\theta=(2k+1)\pi) and purely imaginary when θ=π2+kπ\theta=\frac{\pi}{2}+k\pi. Example: for z=2e−iπ/3z=2\mathrm{e}^{-\mathrm{i}\pi/3}, zn=2ne−inπ/3z^n=2^n\mathrm{e}^{-\mathrm{i}n\pi/3} is real and positive when nπ3=2kπ\frac{n\pi}{3}=2k\pi, so the smallest positive nn is 66, and z6=64z^6=64.

Key termsrealpurely imaginary
Exam tip

Set the argument equal to kπk\pi and solve for the unknown integer or angle.

Section 5

Using the definition in identities

Since 1eiθ=e−iθ=cos⁡θ−isin⁡θ\frac{1}{\mathrm{e}^{\mathrm{i}\theta}}=\mathrm{e}^{-\mathrm{i}\theta}=\cos\theta-\mathrm{i}\sin\theta, adding and subtracting gives, for z=eiθz=\mathrm{e}^{\mathrm{i}\theta}: z+1z=2cos⁡θ,z−1z=2isin⁡θ.z+\frac1z=2\cos\theta,\qquad z-\frac1z=2\mathrm{i}\sin\theta. Squaring gives z2+1z2=(z+1z)2−2=4cos⁡2θ−2z^2+\frac{1}{z^2}=\left(z+\frac1z\right)^2-2=4\cos^2\theta-2. These results let you turn equations in zz into equations in cos⁡θ\cos\theta: for example z2+1z2=−1z^2+\frac{1}{z^2}=-1 becomes cos⁡2θ=14\cos^2\theta=\frac14, giving four values of θ\theta in [0,2π)[0,2\pi).

Key termsreciprocal
Exam tip

When ∣z∣=1|z|=1, 1z=z∗\frac1z=z^*, so z+1z=z+z∗=2 Re(z)z+\frac1z=z+z^*=2\,\mathrm{Re}(z).

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Exam questions on Exponential form of a complex number

  1. The complex number z=2eiπ/3z=2\mathrm{e}^{\mathrm{i}\pi/3}.
    Find z3z^3 in the form a+bia+b\mathrm{i}.2 marks
  2. The complex numbers w1=3eiπ/4w_1=3\mathrm{e}^{\mathrm{i}\pi/4} and w2=2e−iπ/4w_2=2\mathrm{e}^{-\mathrm{i}\pi/4}.
    Express w12w2w_1^2w_2 in the form reiθr\mathrm{e}^{\mathrm{i}\theta}.2 marks
  3. The complex number z=1−3 iz=1-\sqrt3\,\mathrm{i}.
    Express zz in the form reiθr\mathrm{e}^{\mathrm{i}\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).