I'm having trouble evaluating the following kind of integrals around the origin:
$$\oint_{\lvert z =1\rvert} \frac{1}{1-e^z}dz$$
I know how to solve them in the real case (that is, changing "z" to "x" in the integrand) using substitution; however, I'm very confused if I can do such things in the complex case. Also, usually the integrand has exponentials multiplied or divided by $z$, which allows the use of Cauchy's Integral Formula. I tried finding the series representation of $\frac{1}{1-e^z}$, and then finding its residue, but I'd like to know if there is a simpler way of solving this. Any help would be appreciated!
It kinda makes things even more complicated if you simply try to use substitution (i.e. by parameterising the path):
$$\oint_{|z|=1}\frac{dz}{1-e^z} = \int_0^{2\pi}\frac{ie^{it}}{1-\exp(e^{it})}dt$$
which isn't at all easy to integrate.
On the other hand, finding the residues isn't that hard in this case, since the only singularity inside $|z|=1$ is the point $z=0$, where
$$\frac{1}{1-e^z}=\bigl(1-e^z\bigr)^{-1}=-\biggl(z+\frac{z^2}{2}+\cdots \biggr)^{-1}=-\biggl(z^{-1}-\frac 12+\cdots \biggr)$$
So $\text{res}\bigl(\frac{1}{1-e^z} ; 0\bigr)=-1$ and hence the integral equals $-2\pi i$