How to calculate this contour integral?

Question

Calculate $$\oint_C \frac{dz}{(z^2+9)(z+9)}$$ with $C: |z|=4$

I know that the function $\frac{1}{(z^2+9)(z+9)}$ is analytic in $\mathbb{C}$ except in the points $3i,-3i,-9$.

I've tried use the Cauchy's Formula but i don't find the way to use it.

I appreciate your collaboration.

Answer 1

HINT

Given $$\int_c \frac{dz}{(z-3i)(z+3i)(z+9)}$$

The counter is $|z|=4$ so the sigularities in counter are $3i$ and $-3i$ apply cauchy residue theorem

$$\int_c \frac{dz}{(z-3i)(z+3i)(z+9)}=2\pi i \left \{ Res(f,3i)+Res(f,-3i)\right \}$$

Answer 2

Cauchy's residue theorem is the way to go, but if you haven't seen that in your course yet, you can still use Cauchy's integral formula.

Your contour $C$ contains the singularities $\pm3i$, but for CIF you can only deal with one at a time. The trick is to "split" the circular contour $C$ into two smaller contours, each containing one singularity. So let $C_+$ and $C_-$ denote the semicircles, in the upper and lower half plane respectively, centred at $0$ with radius $4$. Then $$\int_C f(z)\,dz=\int_{C_+} f(z)\,dz+\int_{C_-} f(z)\,dz$$ as the integrals over the straight-line segments cancel. Each of these new integrals can be done with CIF: $$\int_{C_+}\frac{dz}{(z^2+9)(z+9)}=\int_{C_+}\frac{g(z)\,dz}{z-3i} =2\pi i\,g(3i)$$ where $$g(z)=\frac{1}{(z+3i)(z+9)}$$ etc.

Answer 3

Here's a more scenic route $$\oint_{|z|=4}\frac{1}{(z^2+9)(z+9)}\ \mathrm dz$$ After some partial fraction decomposition, followed by an application of the estimation lemma, we have $$\frac{1}{90}\oint_{|z|=4}\frac{9-z}{z^2+9}\ \mathrm dz$$ We can further decompose this fraction to again find another contour integral that vanishes. Which leaves us with $$-\frac{1}{180}\oint_{|z|=4}\frac{2z}{z^2+9}\ \mathrm dz$$ $$=-\frac{1}{180}\oint_{\gamma}\frac{1}{w}\ \mathrm dw$$ Where $$\gamma=16e^{2it}+9$$ $$0\leq t\leq2\pi$$ Note that the contour $\gamma$ winds around the origin twice in a counterclockwise direction, which implies that $$-\frac{1}{180}\oint_{\gamma}\frac{1}{w}\ \mathrm dw=-\frac{4\pi i}{180}=-\frac{\pi i}{45}$$ Therefore $$\oint_{|z|=4}\frac{1}{(z^2+9)(z+9)}\ \mathrm dz=-\frac{\pi i}{45}$$