Question

Convert the complex number z=-1-i in to the polar form.

Answer 1

Here $z=-1-i=r(cos\theta +isin\theta )$

$\Rightarrow rcos\theta =-1$ and $rsin\theta =-1\dots i$

Squaring both sides of (i) and adding

$r^2(co{s}^2\theta +si{n}^2\theta )=1+1$

$\Rightarrow r^2=2\Rightarrow r=\sqrt{2}$

$\therefore \sqrt{2}cos\theta =-1$ and $\sqrt{2}sin\theta =-1$

$\Rightarrow cos\theta =\frac{-1}{\sqrt{2}}$ and $sin\theta =\frac{-1}{\sqrt{2}}$

Since $sin\theta$ and $cos\theta$ are both negative

$\therefore \theta$ lies in third quadrant

$\therefore \theta =(-\pi +\frac{\pi }{4})=\frac{-3\pi }{4}$

Hence polar form of z is

$\sqrt{2}[cos\left(\frac{-3\pi }{4}\right)+isin\left(\frac{-3\pi }{4}\right)]$.