Compute integral $\int_0^{2\pi} \ln|R e^{i\phi}-a|d\phi$ How to evaluate this integral without using Jensen's formula?
$\int_0^{2\pi} \ln|R e^{i\phi}-a|d\phi$,
where $R>0$ and $a \in \mathbb{C}$. 
Thank you!
 A: If $a=0$, then he value of the integral is $2\,\pi\ln R$. From now on I will assume $a=\alpha+i\,\beta\ne0$, $\alpha,\beta\in\mathbb{R}$.
First case: $|a|>R$. Then the function
$$
f(x,y)=\frac12\ln\bigl((x-\alpha)^2+(y-\beta)^2\bigr)=\ln|x+i\,y-a|
$$
is harmonic on an open set containing the closed disk $\{(x,y)\in\mathbb{R}^2:x^2+y^2\le R^2\}$. By the mean value property of harmonic functions
$$
\int_0^{2\pi} \ln|R\,e^{i\phi}-a|\,d\phi=2\,\pi\,f(0,0)=\frac12\ln\bigl(\alpha^2+\beta^2\bigr)=2\,\pi\,\ln|a|.
$$
Second case: $|a|&ltR$. Let $a=|a|\,e^{i\psi}$. Then
$$
\int_0^{2\pi} \ln|R\,e^{i\phi}-a|\,d\phi=\int_0^{2\pi} \ln|R\,e^{-i\psi}-|a|e^{-i\phi}|\,d\phi=\int_0^{2\pi} \ln||a|e^{i\phi}-R\,e^{-i\psi}|\,d\phi.
$$
By the result for the first case, this integral is equal to $2\,\pi\ln R$.
Third case: $|a|=R$. Let $a=|a|\,e^{i\psi}$. Then
$$\begin{align*}
\int_0^{2\pi} \ln|R\,e^{i\phi}-a|\,d\phi&=2\,\pi\ln R+\int_0^{2\pi} \ln|e^{i\phi}-e^{i\psi}|\,d\phi\\
&=2\,\pi\ln R+\int_0^{2\pi} \ln|e^{i\phi}-1|\,d\phi\\
&=2\,\pi\ln R.
\end{align*}$$
The las equality follows from
$$\begin{align*}
\int_0^{2\pi} \ln|e^{i\phi}-1|\,d\phi&=\int_0^{\pi} \ln|e^{i\phi}-1|\,d\phi+\int_\pi^{2\pi} \ln|e^{i\phi}-1|\,d\phi\\
&=\int_0^{\pi} \ln|e^{i\phi}-1|\,d\phi+\int_0^{\pi} \ln|e^{i(\phi+\pi)}-1|\,d\phi\\
&=\int_0^{\pi} \ln|e^{i\phi}-1|\,d\phi+\int_0^{\pi} \ln|e^{i\phi}+1|\,d\phi\\
&=\int_0^{\pi} \ln|e^{2i\phi}-1|\,d\phi\\
&=2\int_0^{2\pi} \ln|e^{i\phi}-1|\,d\phi,
\end{align*}$$
which implies
$$
\int_0^{2\pi} \ln|e^{i\phi}-1|\,d\phi=0.
$$
