Solution:

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

Now, $\quad \log (1+\cos \theta+i \sin \theta)=\log \left(1+e^{i \theta}\right)$

Putting, $r=1$ in (i),

$$ \begin{aligned} \log (1+\cos \theta+i \sin \theta) &=\frac{1}{2} \log (1+2 \cos \theta+1)+i \tan ^{-1} \frac{\sin \theta}{1+\cos \theta} \\\\ =& \frac{1}{2} \log [2(1+\cos \theta)]+i \tan ^{-1} \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \\\\ &=\frac{1}{2} \log \left[2 \cdot 2 \cos ^{2} \frac{\theta}{2}\right]+i \tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{1}{2} \log \left[\left(2 \cos \frac{\theta}{2}\right)^{2}\right]+i \frac{\theta}{2} \\\\ &=\frac{1}{2} \cdot 2 \log \left(2 \cos \frac{\theta}{2}\right)+i \cdot \frac{\theta}{2}=\log \left(2 \cos \frac{\theta}{2}\right)+i \cdot \frac{\theta}{2} . \end{aligned}\\ $$