integrate $\frac{1}{(z-e^{-z})}$ on upper half unit circle

Question

I am struggling to integrate $\frac{1}{(z-e^{-z})}$ on upper half unit circle. I would appreciate any support as I am very new to complex integration.

Answer 1

Since you're in the arc $|z| = 1$, and strictly speaking you're interested in the residue inside the unit circle, use Laurent Expansion.

$$\frac{1}{z - e^{-z}} = \frac{1}{e^{-z}(ze^z - 1)} = -\frac{e^z}{1 - ze^z} = -e^z\sum_{k = 0}^{+\infty}z^ke^{zk} = -\sum_{k = 0}^{+\infty}z^ke^{(k+1)z}$$

The series develops as follows

$$-1 - ze^{2z} - z^2e^{3z} - z^3e^{4z}-\ldots$$

By Cauchy formula, the residue is given by

$$2\pi i a_{-1}$$

Where $a_{-1}$ is the coefficient of $z^{-1}$ Laurent Series expansion, which here is missing. Hence the residue is zero and your integral is zero.