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?
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 $$ a_{-1}=\frac{1}{2i\pi}\oint_{C_2}f(w)dw. $$