Proof of Laurent series co-efficients in Complex Residue Am trying to see if there is any proof available for coefficients in Laurent series with regards to Residue in Complex Integration.
The laurent series for a complex function is given by
$$ f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n + \sum_{n=1}^{\infty} \frac{b_n}{(z-z_0)^n} $$
where the principal part co-efficient $$ b_1 = \frac{1}{2 \pi i} \int_C f(z)~dz $$ 
I am unable to understand the proof for $ b_1 $ above.
$ b_1 $ is also called as $ Res_{z=z_0}~f(z)$
http://homepages.math.uic.edu/~jlewis/hon201/laurent.pdf
I was reading the above link for this proof, however could not manage to understand.
Can anyone help please?
 A: Let's first consider $f(z)$ analytic inside and on a boundary $C$ of a simply connected region. Then Cauchy's integral formula is $f(z_0) = \frac{1}{2i\pi}\oint_C\frac{f(z)}{z-z_0}dz$.
Proof: Let $\Gamma$ be a circle of radius $\epsilon > 0$ inside $C$ with center $z_0$. Then $\oint_C\frac{f(z)}{z-z_0}dz = \oint_{\Gamma}\frac{f(z)}{z-z_0}dz$. The equation for a circle with center $z_0$ and radius $\epsilon$ is $\lvert z - z_0\rvert = \epsilon\iff z - z_0 = \epsilon e^{i\theta}$. Let $z = z_0 + \epsilon e^{i\theta}$ so $dz = i\epsilon e^{i\theta}d\theta$.
\begin{align}
\oint_{\Gamma}\frac{f(z)}{z-z_0}dz &= \int_0^{2\pi}\frac{f(z_0 + \epsilon e^{i\theta})}{\epsilon e^{i\theta}}i\epsilon e^{i\theta}d\theta\\
&= i\lim_{\epsilon\to 0}\int_0^{2\pi}f(z_0+\epsilon e^{i\theta})d\theta\\
&= 2i\pi f(z_0)
\end{align}
Thus, $f(z_0) = \frac{1}{2i\pi}\oint_C\frac{f(z)}{z-z_0}dz$.

Let $f(z)$ be analytic in a region bounded by two concentric circles $C_1$ and $C_2$ where the radius of $C_1$ is greater than the radius of $C_2$. By Cauchy's integral formula, we are able to show that $f(z_0) = \frac{1}{2i\pi}\oint_{C_1}\frac{f(z)}{z-z_0}dz-\frac{1}{2i\pi}\oint_{C_2}\frac{f(z)}{z-z_0}dz$.

Laurent's theorem: If $f(z)$ is analytic inside and on the boundary of an annular region bounded by two concentric circles centered at $z_0$ with radii $r_1$ and $r_2$, then for all $z$ in the annular region
$$
f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n+\sum_{n=1}^{\infty}\frac{a_{-n}}{(z-z_0)^n}
$$
where the coefficients are defined as 
\begin{align}
a_n &= \frac{1}{2i\pi}\oint_{C_1}\frac{f(w)}{(w-z)^{n+1}}dw\\
a_{-n} &= \frac{1}{2i\pi}\oint_{C_2}\frac{f(w)}{(w-z)^{-n+1}}dw
\end{align}
Proof: By Cauchy's integral formula, we have
$$
f(z) = \frac{1}{2i\pi}\oint_{C_1}\frac{f(w)}{w-z}dz-\frac{1}{2i\pi}\oint_{C_2}\frac{f(w)}{w-z}dz\tag{1}
$$
for $z$ in the annular region.
Since you are concerned with the $a_{-1}$, I am only going to show the work for the second integral in equation $(1)$. Consider $\frac{-1}{w-z}$. 
\begin{align}
\frac{-1}{w-z} &= \frac{1}{z-z_0}\frac{1}{1 - \frac{w-z_0}{z-z_0}}\\
&= \frac{1}{z-z_0}\sum_{k=0}^{\infty}\Bigl(\frac{w-z_0}{z-z_0}\Bigr)^k\\
&= \frac{1}{z-z_0} + \frac{w-z_0}{(z-z_0)^2} + \frac{(w-z_0)^2}{(z-z_0)^3}+\cdots + \frac{(w-z_0)^{n-1}}{(z-z_0)^n}\\
&+ \text{higher order terms}\\
-\frac{1}{2i\pi}\oint_{C_2}\frac{f(w)}{w-z}dz &= 
\frac{1}{2i\pi}\oint_{C_2}\frac{f(w)}{z-z_0}dw + \frac{1}{2i\pi}\oint_{C_2}\frac{w-z_0}{(z-z_0)^2}f(w)dw + \frac{1}{2i\pi}\oint_{C_2}\frac{(w-z_0)^2}{(z-z_0)^3}f(w)dw\\
&+\cdots + \frac{1}{2i\pi}\oint_{C_2}\frac{(w-z_0)^{n-1}}{(z-z_0)^n}f(w)dw + \text{higher order terms}\\
&= \frac{a_{-1}}{z-z_0} + \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-3}}{(z-z_0)^3}+\cdots + \frac{a_{-n}}{(z-z_0)^n}\\
&+ \text{higher order terms}
\end{align}
Then 
$$
a_{-1}=\frac{1}{2i\pi}\oint_{C_2}f(w)dw.
$$
