Consider the integral
$$I=\int_0^{2\pi}\log\left|re^{it}-a\right|\,dt$$
where $a$ is a complex number and $0<r<|a|$. We have
$$I=\operatorname{Re}\left(\int_0^{2\pi}\log\left|re^{it}-a\right|\,dt\right)$$
Let $\gamma=\partial D(0,r)$. Then
$$\begin{align}\int_\gamma\frac{\log(z-a)}{iz}\,dz&=\int_0^{2\pi}\frac{\log\left(re^{it}-a\right)} {ire^{it}}rie^{it}\,dt\\ &=\int_0^{2\pi}\log\left(re^{it}-a\right)\,dt\end{align}$$
Thus
$$I=\operatorname{Re}\left(\int_{\gamma}\frac{\log(z-a)}{iz}\,dz\right)$$
Now my problem is that $\log(z-a)$ is not holomorphic in $D(0,r)$, so i can't use Cauchy's integral formula to compute $I$. How can I solve this?
Because $0<r<|a|,$ $$\ln(a-z)=\ln{a\left(1-\dfrac{z}{a}\right)}=\ln{a}+\ln{\left(1-\dfrac{z}{a}\right)}=\ln{a}+\sum\limits_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot\left(\frac{z}{a} \right)^k }.$$ Therefore, Laurent expansion for $\frac{\ln(a-z)}{iz}=-i\frac{\ln(a-z)}{z}$ is $$-i\frac{\ln{a}}{z}-i\sum\limits_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot\frac{z^{k-1}}{a^k } }.$$ Using the residue theorem for integral $\int\limits_{\gamma}\frac{\log(z-a)}{iz}\,dz$ gives $$\int\limits_{\gamma}\frac{\log(z-a)}{iz}\,dz=2\pi{i}\cdot(-i\ln{a})=2\pi\ln{a}.$$ Taking the real part, $$\operatorname{Re}\left(\int_{\gamma}\frac{\log(z-a)}{iz}\,dz\right)=\operatorname{Re}(2\pi\ln{a})=2\pi\ln{|a|}.$$