# Contour Integral on Square in complex plane.

If we define $f(z)=\pi \exp(\pi\bar{z})$ and define the contour $\gamma$ to be the boundary of the square with vertices $0$, $1$, $i$ and $1+i$ traversed in the positive direction. I'm trying to compute:

$\displaystyle\int_\gamma f(z)dz$

My thoughts: It seems like a straight-forward application of Cauchy-Goursat. The problem for me is solving it for a square rather than a smooth curve. Any help appreciated.

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It is not a straightforward application of Cauchy, as $f$ is not analytic within $\gamma$. Rather, parametrize each segment of the contour.
\begin{align}\oint_{\gamma} dz \: \pi \, e^{\pi \bar{z}} &= \underbrace{\pi \int_0^1 dx \: e^{\pi x}}_{z=x} + \underbrace{i \pi e^{\pi} \int_0^1 dy \: e^{-i \pi y}}_{z=1+i y} + \underbrace{\pi e^{-i \pi} \int_1^0 dx \: e^{\pi x}}_{z=i+x} +\underbrace{i \pi \int_1^0 dy \: e^{-i \pi y}}_{z=i y}\\ &=\left ( e^{\pi}-1\right) + e^{\pi}\left(1-e^{-i \pi} \right)-e^{-i \pi} \left(e^{\pi}-1 \right )-\left ( 1-e^{-i \pi}\right )\\ &= 4\left ( e^{\pi}-1\right)\end{align}
Note that this is not equal to zero. If you were to replace $\bar{z}$ with $z$ in the original integral and did a similar calculation, you would find that would equal zero.