Calculate all possible values of
$\int_\gamma \frac{\cos(z+1)}{z(z+100i)}dz$ where $\gamma$ is a simple closed curve in C, oriented counter clockwise.
Solution:
The function $\frac{\cos(z+1)}{z(z+100i)}$ has two singularities. $z=0,z=-100i$
- The first possibility is that neither singularity is inside $\gamma$. By Cauchy's Theorem, the integral in this case = 0.
- The second possibility is that $\gamma$ contains exactly one singularity. By choosing simple closed curves around the singularities $z=0, z=-100i$ and applying Cauchy's Integral Formula I found the values of the integral to be $\frac{\pi \cos(1)}{50}$ and $-\frac{\pi \cos(1-100i)}{50}$ respectively.
- The last possibility is that $\gamma$ contains both singularities. This integral is equal to the sum of the integrals in part $2$.
Is that all correct? Do my values in part $2$ look right?