Evaluate
$$ \oint _{|z|=1} \frac{\log\ |1-z|}{z}dz $$ My Attempt $$ I=\oint _{|z|=1} \frac{\log\ |1-z|}{z}dz $$ $$z=e^{i\theta} \Rightarrow dz =i e^{i\theta}d\theta$$ $$I=i \int_{0}^{2\pi} \log\ |1-e^{i\theta}|d\theta$$ $$\text{ define }\, f(z)= |1-z|$$ $f(0)=1$
Now, $f(z)=0$ gives $z=1$.
By Jensen's Formula, $$\frac{1}{2\pi} \int_{0}^{2\pi} \log\ |1-e^{i\theta}|d\theta = \log |f(0)|+\log(1)$$ $f(0)=1$ gives, $$\frac{1}{2\pi} \int_{0}^{2\pi} \log\ |1-e^{i\theta}|d\theta =0$$ $$I=0$$
First of all $\frac{\log\ |1-z|}{z}$ is integrable on $|z|=1$ (it is continuous away from a logarithmic singularity at $z=1$)
$$\int _{|z|=1} \frac{\log\ |1-z|}{z}dz=i \Im(\int _{|z|=1} \frac{\log(1-z)}{z}dz)=\lim_{r\to 1} i\Im(\int _{|z|=r} \frac{\log(1-z)}{z}dz)=0$$ by the Cauchy integral theorem (as $\frac{\log(1-z)}{z}$ is analytic for $|z|<1$)