Define the term $a_n$ in Laurent's series expansion $\sum_{n=-\infty}^{\infty} a_n(z-a)^n$ of $f$ centered on $z=a.$...

Question

Define the term $a_n$ in Laurent's series expansion $\sum_{n=-\infty}^{\infty} a_n(z-a)^n$ of $f$ centered on $z=a.$ classify the singularity $ z=a.$

a)$f(z)=\frac{z^2+1}{z(z+1)}, a=0.$

b)$f(z)=\frac{1}{z(z-1)}$ for $a=0.1.$

we have the following formula $a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz$

So I need to do

a) $a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz = \frac{1}{2 \pi i} \int_{\gamma} \frac{\frac{z^2+1}{z(z+1)}}{z}dz$ ?

b)$a=0 \to a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz = \frac{1}{2 \pi i} \int_{\gamma} \frac{\frac{1}{z(z-1)}}{z}dz$?

$a=1 \to a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz = \frac{1}{2 \pi i} \int_{\gamma} \frac{\frac{1}{z(z-1)}}{z-1}dz$?

If it is that, how do I solve this integral?

Thanks.

Answer 1

You don't need that formula.

a) For each $z\in\Bbb C$ with $|z|<1$, you have\begin{align}\frac{z^2+1}{z+1}&=z-1+\frac2{z+1}\\&=z-1+2\sum_{n=0}^\infty(-1)^nz^n\\&=1-z+2\sum_{n=2}^\infty(-1)^nz^n.\end{align}Therefore, if $z\ne0$,$$\frac{z^2+1}{z(z+1)}=\frac1z-1+2\sum_{n=1}^\infty(-1)^{n+1}z^n.$$So, $0$ is a simple pole.

b) Near $0.1$, $\frac1{z(z-1)}$ is the quotient of two analytic functions, and therefore it is an analytic function. It follows that $0.1$ is a removable singularity of $f$.

Answer 2

For $f(z)=\dfrac 1{z(z-1)}$, we have $f(z)=\dfrac 1{z-1}-\dfrac 1z$.

For $\dfrac 1{z-1}=\dfrac 1{z-0.1-0.9}=-10/9\dfrac 1{1-\dfrac {z-0.1}{0.9}}=\dfrac {-10}9\sum_{n\ge0}(\dfrac {z-0.1}{0.9})^n$, on $\mid z-0.1\mid\lt0.9$.

Similar manipulations can be done to $\dfrac 1z$. $f$ is analytic at $0.1$.