Compute $\int_\gamma \! \frac{e^z-e^{-z}}{e^z-4} \, \mathrm{d}z$ and $\int_\gamma \! \frac{e^z}{z^4} \, \mathrm{d}z$ where $\gamma$ is the path $[A,B,C,D]$ for $A = 1+i$, $B=-1+i$, $C = -1 -i$, $D = 1-i$. (positively oriented rectangle)
I found that the winding number $n_\gamma(0) = 1$ and hence $\int_\gamma \! \frac{1}{z} \, \mathrm{d}x = 2\pi i$ and I should use this to proceed.
Can anyone give me a hint? Thanks.
In the case of the first integral, you can simply use Cauchy's theorem, since the path is null-homotopic in $\mathbb{C}\setminus\{z\in\mathbb{C}\,|\,e^z=4\}=\mathbb{C}\setminus\{\log 4+2k\pi i\,|\,k\in\mathbb Z\}$.
For the other integral, just use the residue theorem. It tells you that$$\int_\gamma\frac{e^z}{z^4}\,\mathrm dz=2\pi i\operatorname{res}_{z=0}\left(\frac{e^z}{z^4}\right).$$