Prove that $\oint \log|z_0 - z| dz = -\frac{1}{2} \oint \frac{z_0 - z}{\bar{z_0} - \bar{z}} d\bar{z}$ I am trying to prove that 
$$\oint \log|z_0 - z| dz = -\frac{1}{2} \oint \frac{z_0 - z}{\bar{z_0} - \bar{z}} d\bar{z}$$
where $z_0$ is a point outside the contour that we're integrating around.
My attempt:
I know that the solution requires an integration by parts, so I tried to integrate the LHS after changing the $\log$ term to $\log|\bar{z_0} - \bar{z}|$ (since modulus is unchanged under conjugation).
LHS = $$\oint \log|\bar{z_0} - \bar{z}| dz$$
Integrate by parts with $u = \log|\bar{z_0} - \bar{z}|$, $v' = 1$, and note that the $[u*v]_{\partial D}$ term is $0$ around a closed contour:
LHS = $$ 0 - \oint \frac{-z}{\bar{z_0} - \bar{z}} dz ?$$
I think that this is incorrect, because $u' \neq \frac{1}{\bar{z_0} - \bar{z}}$, since we need to differentiate with respect to $z$ and not $\bar{z}$.
I know that the extra $z_0$ term in the numerator comes in through the fact that 
$$ \oint \frac{1}{z_0 - z}dz = \oint \frac{1}{\bar{z_0} - \bar{z}}d\bar{z} = 0$$
by the Cauchy Integral Theorem. But I'm just not able to pull all of these ideas together.
Any hints or solutions would be awesome. I've worked really hard to try to understand the proof that this is used in, and this is the last step!

Write
$$\log |z_0-z| = \frac{1}{2} \log |z_0-z|^2.$$
Use chain rule to reduce the derivative with respect to $z$ to
$$ \frac{d}{dz} \left ( \log |z_0-z| \right ) = - \frac{1}{2} \frac{(z_0-z)^* + (z_0-z) \frac{dz^*}{dz}}{(z_0-z)(z_0-z)^*}.$$
Contour integration with respect to $z$ of the first term,
$$ - \frac{1}{2} \frac{1}{z-z_0} $$
yields zero because of Cauchy's theorem (pole $z_0$ lies outside the closed contour). For the second term, 
$$
- \frac{1}{2(z_0-z)^*} \frac{dz^*}{dz}
$$
its $z$-integration
gives, by calculating the residue of its single pole through multiplying by $(z_0-z)$,
$$
\log|z-z_0| = - \frac{(z_0-z)}{2 (z_0-z)^*} \frac{dz^*}{dz}
$$
A further contour integration with respect to $z$ of both sides gives the result.
