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

$\therefore \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4} \quad$ [As $\theta$ lies in the II quadrant]

It can be written,

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

This is the required polar form.