This is regarding a doubt in contour integration [example 1 - at time 4.49 - https://www.youtube.com/watch?v=GnSrxgRjwJU&list=PL6viIjeqW6oNN3wTFQMJ9Vzps1kPzkiF3&index=5]. I did all the problem. I need to find the value of residue. This is found as $\dfrac{\sqrt {-i}}{-2i}$
$-i = e^{\frac{-i\pi}{2}}\tag{1}$
$-i = e^{\frac{i3\pi}{2}}\tag{2}$
Using $(1)$, $\text{value of residue} = \dfrac{e^{\frac{-i\pi}{4}}}{-2i} = \dfrac{\sqrt 2}{4}+\dfrac{\sqrt 2}{4}i\tag{3}$
Using $(2)$, $\text{value of residue} = \dfrac{e^{\frac{i3\pi}{4}}}{-2i} = -\dfrac{\sqrt 2}{4}-\dfrac{\sqrt 2}{4}i\tag{4}$
Due to this, my residue sum changed. But when I saw the solution of problem, he took $-i$ as $(2)$ instead of $(1)$
But $-i$ is negative $y-$axis. The answer should come same. But it is differing. I am not able to understand here. Pls, explain.
He has taken $-i$ as $(2)$ because of where his branch cut is placed.
In order to calculate the real integral $\int^\infty_0f(x)dx$ he has used what he calls the "Pac Man contour". This contour places a cut along the positive $x$-axis. He has effectively restricted himself to using $\{0<\mathrm{arg}(z)<2\pi\}$.
If his "Pac Man contour" had been the other way around, with the cut along the negative $x$-axis (in order, for example, to calculate $\int_{-\infty}^0f(x)dx$) then he would use $\{-\pi<\mathrm{arg}(z)<\pi\}$ and would take $-i$ as $(1)$.
I would recommend reading more about branch cuts and why they are necessary.