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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.
In addition to José Carlos Santos' answer, we can also utilize Green's theorem:
\begin{align*} \int_{\gamma} \pi e^{\pi \bar{z}} \, dz &= \int_{\gamma} \left( \pi e^{\pi \bar{z}} \, dx + i\pi e^{\pi \bar{z}} \, dy \right) \\ &= \int_{[0,1]^2} \left( \frac{\partial}{\partial x} i\pi e^{\pi \bar{z}} - \frac{\partial}{\partial y} \pi e^{\pi \bar{z}}\right) \, dxdy \\ &= 2\pi^2 i \int_{[0,1]^2} e^{\pi x}e^{-i\pi y} \, dxdy \\ &= 4(e^{\pi} - 1). \end{align*}
You can't use the residue theorem in a non-analytic function!
Your integral can be naturally broken on four pieces. One of them is$$\int_\gamma\pi e^{\pi\overline z}\,\mathrm dz,\tag{1}$$with $\gamma\colon[0,1]\longrightarrow\mathbb C$ defined by $\gamma(t)=t$. But then $(1)$ is equal to$$\int_0^1\pi e^{\pi\overline{\gamma(t)}}\gamma'(t)\,\mathrm dt=\int_0^1\pi e^{\pi t}\,\mathrm dt=\left[e^{\pi t}\right]_{t=0}^{t=1}=e^{\pi}-e^0=e^{\pi}-1$$Can you compute the other three integrals?
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 $$\int_{\gamma}\pi e^{\pi\overline{z}}dz=\int_{t=0}^1\pi e^{\pi(\overline{P}+(\overline{Q}-\overline{P})t)}(Q-P)dt=\frac{Q-P}{\overline{Q}-\overline{P}}[e^{\pi(\overline{P}+(\overline{Q}-\bar{P})t)}]_0^1=\frac{(Q-P)(e^{\pi \overline{Q}}-e^{\pi \overline{P}})}{\overline{Q-P}}.$$
