Question

Convert the given complex number in polar form : $-1+i$

Answer 1

Given, $z=-(1-i)$
Let $r\cos \theta =1$ and $r\sin \theta =-1$
On squaring and adding, we obtain
$r^2{\cos }^2\theta +r^2{\sin }^2\theta =1^2+{(-1)}^2$
$\Rightarrow r^2({\cos }^2\theta +{\sin }^2\theta )=1+1$
$\Rightarrow r^2=2$
$\Rightarrow r=\sqrt{2}$ (Conventionally r>0 )
$\therefore \sqrt{2}\cos \theta =1$ and $\sqrt{2}\sin \theta =-1$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$ and $\sin \theta =-\frac{1}{\sqrt{2}}$
$\therefore \theta =-\frac{\pi }{4}$ [As $\theta$ lies in the Ii quadrant]
So, the polar form of $z=-(1-i)$ is
$\therefore -(1-i)=-(r\cos \theta +ir\sin \theta )=-\left(\sqrt{2}\cos (-\frac{\pi }{4})+i\sqrt{2}\sin (-\frac{\pi }{4})\right)$
$=-\sqrt{2}[-\cos \left(\frac{3\pi }{4}\right)-i\sin \left(\frac{3\pi }{4}\right)]$
$=\sqrt{2}\left[\cos \left(\frac{3\pi }{4}\right)+i\sin \left(\frac{3\pi }{4}\right)\right]$