complex analysis is life - 4

[logistic_guy]

Solve.

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

where \(\displaystyle C\) is the path that goes from \(\displaystyle i\) to \(\displaystyle 0\).

 

[logistic_guy]

We have:

\(\displaystyle z = x + iy\)
\(\displaystyle \overline{z} = x - iy\)

By observing the path, we see that \(\displaystyle x = 0\), then we have:

\(\displaystyle z = iy\)
\(\displaystyle \overline{z} = -iy\)

This gives us the integral:

\(\displaystyle \large \int_{1}^{0} \pi e^{-i\pi y} \ i dy\)

 

[logistic_guy]

\(\displaystyle \large \int_{1}^{0} \pi e^{-i\pi y} \ i dy\)

Let us solve this bastard integral.

\(\displaystyle \large i\pi\int_{1}^{0} e^{-i\pi y} \ dy = -e^{-i\pi y}\bigg |_{1}^{0} = -(1 - e^{-i\pi}) = e^{-i\pi} - 1 = \cos \pi - i\sin \pi - 1 = -1 - 1 = -2\)

 

[khansaheb]

... bastard.....

Language...Language