Question

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

Answer 1

Given complex number is $-1-i$

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

$\theta =-(\pi -\frac{\pi }{4})=-\frac{3\pi }{4}$ [ As $\theta$ lies in the III quadrant ]

$\therefore -1-i=r\cos \theta +ir\sin \theta =\sqrt{2}\cos \frac{-3\pi }{4}+i\sqrt{2}\sin \frac{-3\pi }{4}$

$=\sqrt{2}(\cos \frac{-3\pi }{4}+i\sin \frac{-3\pi }{4})$

This is the required polar form.