Question

Convert each of the complex numbers in he polar form:

$1-i$

Answer 1

Here $z=1-i=4(cos\theta +isin\theta )$

$\Rightarrow rcos\theta =1$ and $rsin\theta =-1\dots \left(i\right)$

Squaring both sies 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 $sin\theta$ is negative and $cos\theta$ is positive

$\therefore \theta$ lies in fourth quadrant

$\therefore \theta =\frac{-\pi }{4}$

Hence polar form of z is

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