Convert the given complex number in polar form: 1 – i

$1-i$

Let $r \cos \theta=1$ and $r \sin \theta=-1$

On squaring and adding, we obtain

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=1^{2}+(-1)^{2}$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+1$

$\Rightarrow r^{2}=2$

$\Rightarrow r=\sqrt{2} \quad[$ 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 IV quadrant]

$\therefore 1-i=r \cos \theta+i r \sin \theta=\sqrt{2} \cos \left(-\frac{\pi}{4}\right)+i \sqrt{2} \sin \left(-\frac{\pi}{4}\right)=\sqrt{2}\left[\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right]$ This is the required polar form.