Polar Form of a Complex Number

Polar Form of a Complex Number

Let the complex number $z=a+ib$ be represented by the point $P(a,b)$ (Fig. 1.4). Let $m\angle XOP=\theta=\tan^{-1}\left(\dfrac{b}{a}\right)$ (quadrant-corrected as in Section 1.5.3) and $l(OP)=r=\sqrt{a^2+b^2}>0$. Then $P(r,\theta)$ are called the polar coordinates of $P$, and the origin is called the pole.

Since $a=r\cos\theta$ and $b=r\sin\theta$ (Section 1.5.2), $z=a+ib$ becomes

This is called the polar form of the complex number $z=a+ib$.

Worked Example: represent $4+4\sqrt3i,\ -2,\ 3i,\ -\sqrt3+i$ in polar form.

1. $z=4+4\sqrt3i$: $a=4,b=4\sqrt3$, so $r=\sqrt{16+48}=\sqrt{64}=8$. Since $\theta$ lies in Quadrant I: $\theta=\tan^{-1}\left(\dfrac{4\sqrt3}{4}\right)=\tan^{-1}(\sqrt3)=\dfrac{\pi}{3}$ ($60°$). So the polar form is $z=8(\cos60°+i\sin60°)=8\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)$.
2. $z=-2$: $a=-2,b=0$, so $r=\sqrt{4+0}=2$. Since $(-2,0)$ lies on the negative real axis, $\theta=\pi$ ($180°$). Polar form: $z=2(\cos\pi+i\sin\pi)$.
3. $z=3i$: $a=0,b=3$, so $r=\sqrt{0+9}=3$. Since $(0,3)$ lies on the positive imaginary axis, $\theta=\dfrac{\pi}{2}$ ($90°$). Polar form: $z=3\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)$. …