We are being tested next week on the Cauchy Integral Formula, and were given these questions as guidelines.
1) $\int_\gamma \frac{\cos(z)dz}{z(z^2+1)}$ where $\gamma$ is the circle with center $0$ and radius $3$
2) $\int_\gamma \frac{e^z dz}{z^2+1}$ where $\gamma$ is the circle centered at $i$ with radius $1$
3) $\int_\gamma \frac{zdz}{z^2+4}$ where $\gamma$ is the rectangle with vertices $1 \pm 4i$
4) $\int_\gamma \frac{dz}{z^3+1}$ where $\gamma$ is the circle with center $-1$ and radius $\frac{7}{4}$
5) $\int_\gamma \frac{dz}{z^3+1}$ where $\gamma$ is the circle with center $0$ and radius $4$
I solved question 2 by doing the following:
Solved for the poles; $z = \pm i$
Only the pole at i is inside $\gamma$, so write the function as $\frac{\frac{e^z}{z+i}}{z-i}$, which is equal to $2\pi i \cdot \frac{e^i}{2i}$, which gives $\pi(\cos(1)+i\sin(1))$
With regards to question 3, $\gamma$ confused me, as I had a hard time telling what poles were inside it.
With the other questions, I was unsure of how to re-write the function in terms of poles when there were more than 2 poles.
For example, for question 1, I found the poles to be $z = 0$, $z = \pm i$ This could be re-written as $\frac{\frac{\frac{cos(z)}{z}}{z-i}}{z+i}$, but I am unsure of how to proceed. Should I let $z = -i$ and sub that into $\frac{\frac{\cos(z)}{z}}{z-i}$?
I was having the same issue with 4 and 5 where I didn't know how to proceed when the function was re-written with 3 poles.