Exercise 1.4 — Question 10

Polar Representation of Complex Numbers

A complex number $z=a+ib$ can be written in polar (trigonometric) form as:

$z=r(\cos \theta +i\sin \theta )$

where:

• $r=|z|=\sqrt{a^2+b^2}$ is the modulus (distance from origin)
• $\theta =\arg (z)={\tan }^{-1}\left(\frac{b}{a}\right)$ is the argument (angle with positive real axis)
• $a=r\cos \theta$, $b=r\sin \theta$

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Operations in Polar Form

Multiplication

If $z_1=r_1(\cos {\theta }_1+i\sin {\theta }_1)$ and $z_2=r_2(\cos {\theta }_2+i\sin {\theta }_2)$, then:

$z_1\cdot z_2=r_1{r}_2[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)]$

Rule: Multiply the moduli and add the arguments.

Division

$\frac{z_1}{z_2}=\frac{r_1}{r_2}[\cos ({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2)],r_2\ne 0$

Rule: Divide the moduli and subtract the arguments.

De Moivre's Theorem

For any integer $n$:

$z^n=r^n(\cos n\theta +i\sin n\theta )$

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Worked Example

Express $z_1=1+i$ and $z_2=\sqrt{3}-i$ in polar form, then find $z_1{z}_2$ and $\frac{z_1}{z_2}$.

Step 1 — Polar form of $z_1=1+i$:

$r_1=\sqrt{1^2+1^2}=\sqrt{2},{\theta }_1={\tan }^{-1}\left(\frac{1}{1}\right)=45°$

$z_1=\sqrt{2}(\cos 45°+i\sin 45°)$

Step 2 — Polar form of $z_2=\sqrt{3}-i$:

$r_2=\sqrt{(\sqrt{3}{)}^2+(-1{)}^2}=\sqrt{3+1}=2,{\theta }_2={\tan }^{-1}\left(\frac{-1}{\sqrt{3}}\right)=-30°$

$z_2=2(\cos (-30°)+i\sin (-30°))$

Step 3 — Product $z_1{z}_2$:

$z_1{z}_2=\sqrt{2}\cdot 2[\cos (45°+(-30°))+i\sin (45°+(-30°))]$ $=2\sqrt{2}(\cos 15°+i\sin 15°)$

Step 4 — Quotient $\frac{z_1}{z_2}$:

$\frac{z_1}{z_2}=\frac{\sqrt{2}}{2}[\cos (45°-(-30°))+i\sin (45°-(-30°))]$ $=\frac{\sqrt{2}}{2}(\cos 75°+i\sin 75°)$

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