Calculate the complex integral

Question

I have $$ \int{\frac{dz}{z^2+9}} $$ Also I'm given 2 different conditions. First is $|z|=\pi$, second is $|z-2i|=2$.

Okay, so for the integral i have $\int{\frac{dz}{(z+3i)(z-3i)}}$.

For the first condition, if I draw a circle, then $\pi$ will be outside the circle so I can't continue integral calculation using Cauchy's theorem. And for the second one, I'm again not sure how to proceed.

Please shed some light here. Thanks.

Answer 1

For $\mid z\mid=\pi$ the poles are both inside the circle. For $\mid z-2i\mid =2$ only $3i$ is within the region.

Now use the residue theorem. That is, $f(z)=\frac1{z^2+9}=\frac1{6i}(\frac1{z-3i}-\frac1{z+3i})$. The residue at $3i$ is $\frac1{6i}$. So the second integral is $2\pi i\cdot\frac1{6i}=\frac{\pi}3$.

As for the first, we get the difference of the integrals over two smaller contours, both of which are $\frac{\pi}3$, so $0$.

Answer 2

For $|z|=\pi$ the residue theorem implies that $\int \frac{1}{(z+3i)(z-3i)}dz = 2\pi i [(Res f, 3i) + (Res f, -3i)]$ where $f(z) = \frac{1}{z^2+9}$. The other case is similar but only one of the poles lies in the interior of the circle.