TITLE: Complex integral with conjugate as an exponent
QUESTION [7 upvotes]: What is the integral of 
  $$\int_{\Gamma}\pi e^{\pi\bar{z}}dz$$ where $\Gamma$ is the square with vertices at $0,1,1+i,i$ oriented anticlockwise? 

I am badly stuck at this problem. I thought of using the Residue theorem by using $\bar{z}=\frac{|z|^2}{z}$, but we get an essential singularity. Using Laurent series about zero, I get $a_{-1}=\pi|z|^2$. Is this correct? how do we proceed? Any hints. Thanks beforehand.

REPLY [3 votes]: Hint. Note that if $\gamma$ is a segment from $P$ to $Q$ then
$\gamma(t)=P+(Q-P)t$ for $t\in [0,1]$ and
$$\int_{\gamma}\pi e^{\pi\overline{z}}dz=\int_{t=0}^1\pi e^{\pi(\overline{P}+(\overline{Q}-\overline{P})t)}(Q-P)dt=\frac{Q-P}{\overline{Q}-\overline{P}}[e^{\pi(\overline{P}+(\overline{Q}-\bar{P})t)}]_0^1=\frac{(Q-P)(e^{\pi \overline{Q}}-e^{\pi \overline{P}})}{\overline{Q-P}}.$$