I have $$\int_{\gamma} {\frac{z}{\overline z}}dz$$ where $\gamma$ is the edge of $\{1 < |z| < 2\ $and $\Im (z) > 0\}$.
I think the way to solve this is to calculate the integral for $|z|=1$ and then $|z|=2$, then subtract the first result from the second one.
For $|z|=1$, the integral is $i\pi$. For $|z|=1$, the integral is $4i$. So my answer is $4i-i\pi$. But the answer in book is ${\frac{4}{3}}$ .
Actually, in the upper semi-circle $\lvert z\rvert=1$, you have$$\frac z{\overline z}=\frac{z^2}{\lvert z\rvert^2}=z^2$$and therefore the integral (if you move along the semi-circle from the left to the right) is equal to $\dfrac23$. By the same argument, the integral along the larger semicircle (this time moving from the right to left) is $-\dfrac43$.
Can you take it from here?