Question

Write (i) $-1-i$ (ii) $1-i$ in polar form.

Answer 1

A complex number $a+ib$ in polar form is written as

$r\cos \theta +ir\sin \theta$

So,

$\cos \theta =\frac{a}{r}$

$\sin \theta =\frac{b}{r}$

Hence, $r=\sqrt{a^2+b^2}$

$Part\left(i\right)$

$-1-i$

$a=-1,b=-1$

So, $r=\sqrt{(-1{)}^2+(-1{)}^2}$

$r=\sqrt{2}$

And hence

$\cos \theta =\frac{-1}{\sqrt{2}}\Rightarrow \theta =\frac{3\pi }{4}$

Therefore, the polar form of $-1-i$ is

$=(\sqrt{2},\frac{3\pi }{4})$

$Part\left(ii\right)$

$1-i$

$a=1,b=-1$

So, $r=\sqrt{(1{)}^2+(-1{)}^2}$

$r=\sqrt{2}$

And hence

$\cos \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4}$

Therefore, the polar form of $1-i$ is

$=(\sqrt{2},\frac{\pi }{4})$

Hence, this is the answer.