Write 1 − i in polar form.

$z=1-i$

$r=|z|$

$=\sqrt{1+1}$

$=\sqrt{2}$

Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

$\therefore \tan \alpha=\left|\frac{-1}{1}\right|$

$=\frac{\pi}{4}$

$\Rightarrow \alpha=\frac{\pi}{4}$

Since point $(1,-1)$ lies in the fourth quadrant, the argument of $z$ is given by

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

Polar form $=r(\cos \theta+i \sin \theta)$

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

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