Evaluate a complex integral using Cauchy formula Evaluate 
$$\int_\gamma \frac{\overline{w-z}}{w-z} dw,$$
where where $\gamma$ is the unit circle traversed once in the counterclockwise direction, and z is a point inside the unit disk.
I tried rewriting it into 
$$\int_\gamma \frac{|{w-z}|^2}{{(w-z)}^2} dw,$$
But I'm not sure how that will help.
 A: Hint: Since $\gamma$ is the unit circle and $w$ is on $\gamma$ you have 
$$\bar{w}=\frac{1}{w}$$
Your integral is then
$$\int_\gamma \frac{\overline{w-z}}{w-z} dw,=\int_\gamma \frac{1}{w(w-z)} dw - \bar{z}\int_\gamma \frac{1}{(w-z)} dw,$$
Now the second integral is known, while the first is done either by Partial fraction decomposition, or by putting small nonintersecting curves around $0$ and $z$. Note hat with this approach you should disscuss the case $z=0$ separatelly...
A: You can work with your rewriting: On $\gamma$, we have
$$
\frac{|w-z|^2}{(w-z)^2} = \frac{|w|^2 - w\bar z - \bar w z + |z|^2}{(w-z)^2} = \frac{1 + w\bar z + w^{-1} z + |z|^2}{(w-z)^2}
$$
since $|w|=1$ along the curve. Split into four integrals.
The first and the fourth vanish (why?) and you are left with
$$
\int_\gamma \frac{w\bar z}{(w-z)^2}\,dw = 2\pi i\bar z
$$
and
$$
\int_\gamma \frac{z}{w(w-z)^2}\,dw = 2\pi i \big( \frac1z - \frac1z \big) = 0
$$
by some favourite combination of Cauchy's integral formula, the residue theorem and/or partial fractions decomposition.
