I know the basics of the poles and the residues. However I am stuck at a point. I am trying to understand the fact that for the following integral
$\int_{\gamma R} \dfrac{(e^{az})}{ (1+e^z)} dz $
Where $\gamma R$ is a closed curve formed by the points $R,-R,R+2i\pi , -R+2i\pi$, has simple pole at $z=i\pi$.
The step that was done for proving the pole is simple, was as below...
Since $(1+{e^{z}})'=e^{z} \neq 0$ for $z=i\pi$ hence th pole is a simple pole.
I didn't read any definition of this type. Will anybody please explain?
Thank You
A point $z$ is called a pole of $f(z)$ if it is a zero of some order $m$ of $\frac{1}{f(z)}$.
So here if $f(z)=\frac{1}{1+e^{z}}$ then what is the order of the zero of $1+e^{z}$ at $z=i\pi$ ?.
Well say it is of order $m$.
Then $\frac{1}{f(z)}=(z-i\pi)^{m}g(z)$ where $g(z)$ is a holomorphic function which is non zero at $z=i\pi$.
Then $1+e^{z}=(z-i\pi)^{m}g(z)$.
Differentiatiing :-
$e^{z}=m(z-i\pi)^{m-1}g(z)+(z-i\pi)^{m}g'(z)$
But if you plug in $z=i\pi$ the right hand side is non-zero but the left hand side is zero for $m\geq 2$. So the order $m$ must be $1$.
Hence they are concluding so.
So you have $i\pi$ is a simple pole.