Question

Convert the complex numbers in to the polar form:

-i +i

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 $sin\theta$ is positive and $cos\theta$ is negative

$\therefore \theta$ lies in second quadrant

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

Hence polar form of z is

$\sqrt{2}(cos\frac{3\pi }{4}+isin\frac{3\pi }{4})$.