Solution:

Let, $1+i=r(\cos \theta+i \sin \theta)=r \cos \theta+i \sin \theta$ Equating real \& imaginary parts on both sides,

$$r \cos \theta=1 \quad \& \quad r \sin \theta=1\\ $$

Now,

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

Now,

$(\operatorname{rcos} \theta)^{2}+(\operatorname{rsin} \theta)^{2}=(1)^{2}+(1)^{2} \\ $

$ \Rightarrow \mathrm{r}^{2} \cos ^{2} \theta+\mathrm{r}^{2} …