Euler's relation and polar formEdexcel International A Level Further Maths: Revision notes

Section 1

Euler's relation

For real $\theta$ , Euler's relation states $e^{i\theta}=\cos\theta+i\sin\theta.$ So $e^{i\theta}$ has modulus $1$ and argument $\theta$ . Useful special cases: $e^{i\pi}=-1$ (so $e^{i\pi}+1=0$ ), $e^{i\pi/2}=i$ and $e^{2\pi i}=1$ . Replacing $\theta$ by $-\theta$ gives $e^{-i\theta}=\cos\theta-i\sin\theta$ , the complex conjugate of $e^{i\theta}$ . The exponent $\theta$ is in radians.

Key termsEuler's relationconjugate

Common mistake

Using degrees in the exponent. $e^{i\theta}$ needs $\theta$ in radians.

Section 2

Exponential (polar) form

A complex number with modulus $r$ and argument $\theta$ can be written $z=r(\cos\theta+i\sin\theta)=re^{i\theta}.$ To convert $z=x+iy$ : $r=\sqrt{x^2+y^2}$ , and $\theta$ from $\tan^{-1}\left|\frac yx\right|$ adjusted for the quadrant, with $-\pi<\theta\le\pi$ for the principal argument. For example $z=-1+i\sqrt3$ is in the second quadrant, so $\theta=\pi-\frac\pi3=\frac{2\pi}{3}$ and $z=2e^{2\pi i/3}$ . To go back, $2e^{i\pi/3}=2\left(\frac12+\frac{\sqrt3}{2}i\right)=1+i\sqrt3$ .

Key termsexponential formprincipal argument

Common mistake

Taking $\tan^{-1}\frac yx$ without checking the quadrant. $-1+i\sqrt3$ is not $2e^{-i\pi/3}$ .

Section 3

Multiplying and dividing

The index laws give the rules you already know for modulus and argument: $re^{i\alpha}\times se^{i\beta}=rs\,e^{i(\alpha+\beta)},\qquad \frac{re^{i\alpha}}{se^{i\beta}}=\frac rs\,e^{i(\alpha-\beta)}.$ Multiply the moduli and add the arguments; divide the moduli and subtract the arguments. Example: $2e^{i\pi/3}\times3e^{i\pi/6}=6e^{i\pi/2}=6i$ . If the new argument is outside $(-\pi,\pi]$ , add or subtract $2\pi$ .

Key termsmodulusargument

Exam tip

To add or subtract complex numbers, convert back to $a+ib$ first; the exponential form is for products and quotients.

Section 4

Cosine and sine in exponential form

Adding and subtracting $e^{i\theta}=\cos\theta+i\sin\theta$ and $e^{-i\theta}=\cos\theta-i\sin\theta$ gives $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2},\qquad \sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}.$ With $z=e^{i\theta}$ these read $z+\frac1z=2\cos\theta$ and $z-\frac1z=2i\sin\theta$ . They are very useful for proving identities.

Key termsexponential form of cosine and sine

Common mistake

Forgetting the $i$ in the denominator for sine: $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ .

Section 5

Using the forms to prove identities

Substitute the exponential forms, expand, then convert back. Example: $\left(z+\frac1z\right)^2=z^2+2+\frac1{z^2}$ , so $4\cos^2\theta=2\cos2\theta+2$ , giving $\cos^2\theta=\frac12(1+\cos2\theta)$ . Example: sum-to-product. Writing $A=p+q$ and $B=p-q$ gives $e^{iA}+e^{iB}=e^{ip}(e^{iq}+e^{-iq})$ and hence $\cos A+\cos B=2\cos p\cos q$ with $p=\frac{A+B}{2}$ , $q=\frac{A-B}{2}$ . Use such results to solve equations: $\cos\theta+\cos3\theta=2\cos2\theta\cos\theta$ .

Exam tip

Group terms so that each bracket becomes $e^{ik}+e^{-ik}$ or $e^{ik}-e^{-ik}$ , which you can replace by $2\cos k$ or $2i\sin k$ .

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Euler's relation and polar form

Let $z=2e^{i\pi/3}$ and $w=3e^{i\pi/6}$ .
Find $z+w$ in the form $a+ib$ , giving $a$ and $b$ as exact values.2 marks
Let $\theta$ be a real number and let $z=e^{i\theta}$ .
By expanding $\left(z+\dfrac1z\right)^2$ and using $z^2=e^{2i\theta}$ , show that $\cos^2\theta=\frac12(1+\cos2\theta)$ .2 marks
The complex number $z=-1+i\sqrt3$ .
Write $z$ in the form $re^{i\theta}$ , where $-\pi<\theta\le\pi$ .3 marks

See the full worksheet

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).