Solution:

Deduce that, $\log (1+\cos \theta+i \sin \theta)=\log \left(2 \cos \frac{\theta}{2}\right)+i \frac{\theta}{2}\\$. $\log \left(1+r e^{i \theta}\right)=\log [1+r(\cos \theta+i \sin \theta)]=\log [(1+r \cos \theta)+i(r \sin \theta)]$ $$ \begin{aligned} &=\frac{1}{2} \log \left[(1+r \cos \theta)^{2}+(r \sin \theta)^{2}\right]+i \tan ^{-1} \frac{r \sin \theta}{1+r \cos \theta} \\ &=\frac{1}{2} \log \left[1+2 r \cos \theta+r^{2} \cos ^{2} …