Write z = (1 – i) in polar form.

We have, z = (1 – i)

Let 1 = rcosθ and -1 = rsinθ

By squaring and adding, we get

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

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

$\Rightarrow 2=r^{2}$

$\Rightarrow r=\sqrt{2}$

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

Since, θ lies in fourth quadrant, we have

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

Thus, the required polar form is $\sqrt{2}\left(\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right)$