Calculate $\oint_\gamma \frac{1}{z(z^2+4)} dz$ over a circle

Question

The task I'm given is to evaluate $\displaystyle \oint_\gamma \frac{1}{z(z^2+4)} dz$, where $\gamma$ is the circle with radius $1$ and center point $2i$. I tried with Cauchy's Integral Theorem, but struggled so far.

Answer 1

Hint $$\frac{1}{z(z^2+4)}=\frac{\frac{1}{z(z+2i)}}{z-2i}$$

Set $g(z)=\frac{1}{z(z+2i)}$ in Cauchy Integral Formula.

Answer 2

Step 1: Calculate the locations of the poles. This is where the denominator of the integrand vanishes. Since $$ \frac{1}{z(z^2+4)}=\frac{1}{z}\cdot\frac{1}{z-2i}\cdot\frac{1}{z+2i}, $$ we see that $0,\pm 2i$ are the poles.

Step 2: Determine which locations of the poles are within the circle $\gamma$, in this case, $0$ is of distance $2$, $-2i$ is of distance $4$, so only $2i$ is within the circle.

Step 3: Calculate the residues, at $z=2i$, see this function is $$ \frac{1}{z-2i}\left(\frac{1}{z}\cdot\frac{1}{z+2i}\right)$$ We see that $z-2i$ vanishes to order $1$ at $z=2i$ and the rest is holomorphic near $z=2i$, so we can just plug in to get the residue. In particular, the residue is $$ \frac{1}{2i}\cdot\frac{1}{2i+2i}=-\frac{1}{8}. $$

Step 4: Apply Cauchy's integral formula to get that the integral is $2\pi i\left(-\frac{1}{8}\right)=\frac{-\pi i}{4}$.