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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.

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  • $\begingroup$ Start by writing $z= e^{i\theta}$ $\endgroup$ – user247327 Feb 24 '16 at 13:15
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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.

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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...

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