I have a function expanded as Laurent Series $f(z)=\sum_{n=-\infty}^{\infty}c_n(z-z_0)^n$
It says that for some $r$, $r<|z-z_0|<\infty$, $c_{-1}$ is the sum of the Residues of $f(z)$ at all singularities in the finite part of the plane and that $-c_{-1}$ is the residue of $f(z)$ at $z_{\infty}.$
So somehow it says for a $r$, $-c_{-1}$ is equal to sum of all residues. How can it be possible? I have no idea where to start to prove this?
Here I tried something which is,
$\oint f(z)=\sum_{n=0}^{\infty}a_n\oint (z-z_0)dz+\oint \frac{b_n}{z-z_0}dz+\sum_{n=2}^{\infty}\oint \frac{b_n}{(z-z_0)} dz$
We can write this equality for our case as $\oint f(z)=\sum_{n=0}^{\infty}c_n\oint (z-z_0)dz+\oint \frac{c_{-1}}{z-z_0}dz+\sum_{n=2}^{\infty}\oint \frac{c_{-n}}{(z-z_0)} dz$
Also I suggested that $Res(\frac{c_{-1}}{(z-z_0)^n})=0$ by residue thm and $Res(c_{-1}{(z-z_0)^n})=0$ since its limit goes to 0
So finally $\oint f(z)=2\pi i\sum Res(f(z),z_0)= \oint \frac{c_{-n}}{z-z_0}dz = 2 \pi i \lim_{z \to z_0}(z-z_0)(\frac{c_{-1}}{z-z_0})={c_-1}$
$\oint f(z)={c_-1}$
The problem is asking to prove that it is equal to $-c_{-1}$
Am I right?