complex analysis is life - 2

[logistic_guy]

Solve.

$\displaystyle \large \int_C \pi e^{\pi \overline{z}} \ dz$

where $\displaystyle C$ is the path that goes from $\displaystyle 1$ to $\displaystyle 1 + i$.

 

[logistic_guy]

In the previous exercise we have seen that $\displaystyle z = x + iy$ and $\displaystyle \overline{z} = x - iy$.

The path tells us that $\displaystyle x = 1$.

Then, we have:

$\displaystyle z = 1 + iy$

And

$\displaystyle \overline{z} = 1 - iy$

This reduces our integral to:

$\displaystyle \large \int_0^{1} \pi e^{\pi(1 - iy)} \ idy$

 

[logistic_guy]

$\displaystyle \large \int_0^{1} \pi e^{\pi(1 - iy)} \ idy$

Let us try to solve this nasty integral.

$\displaystyle \large \int_0^{1} \pi e^{\pi(1 - iy)} \ idy = i\pi e^{\pi}\int_{0}^{1} e^{-i\pi y} \ dy = -e^{\pi}e^{-i\pi y}\bigg |_{0}^{1}$


$\displaystyle = -e^{\pi}(e^{-i\pi} - 1) = -e^{\pi}(\cos \pi - i\sin \pi - 1) = -e^{\pi}(-2) = 2e^{\pi}$