Using Local Cauchy Theorem, find the integrals $\int_{|z|=1} \sqrt{9-z^2}\, dz$ and $\int_{|z|=1}(z^2+2z)^{-1}\, dz$.

Question

Using Local Cauchy Theorem, find the integrals $\int_{|z|=1} \sqrt{9-z^2}\, dz$ and $\int_{|z|=1}(z^2+2z)^{-1}\, dz$.

For the first integral, I have no idea what to do.

For the second integral partial fractions gives $$\frac{1}{(z^2+2z)}=\frac{1}{2}\left(\frac{1}{z}-\frac{1}{z+2}\right)$$

But then how do I integrate $\int_{|z|=1} \frac{1}{z+2}$?

Answer 1

If, in your first integral, $\sqrt{9-z^2}$ is a holomorphic square root of $9-z^2$ defined on $\Bbb C\setminus\bigl((-\infty,-3]\cup[3,\infty)\bigr)$, then then answer is $0$, since the while close disk centered at $0$ with radius $1$ is contained in the domain of the function.

For the same reason $\int_{|z|=1}\frac{\mathrm dz}{z+2}=0$.