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.

• Start by writing $z= e^{i\theta}$ Feb 24, 2016 at 13:15

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...
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.