Finding Laurent Series-is it possible without contour integral

Question

I have to find the Laurent Series expansion for $$ f(z)= \frac {z^2 +1} {2z-1} $$

around $ 1 \over 2$. I know that I can write $f(z)= \sum_{-\infty}^{+\infty} a_n(z- \frac{1}{2})^n $ , where $a_n = \frac{1}{2πi} \oint_γ \frac{f(z)dz}{(z-\frac{1}{2})^{n+1}} $ (γ is a closed path around 1/2).

Can I find the series expansion in a more "crude" way, using the geometric series? Can I split the fraction in such a way to form something like $1\over z-1$ , or do I have to use the contour integral?

Moreover, in rational functions like f, is there always a way for building the expansion using the geometric series?

Answer 1

An idea:

$$\frac{z^2+1}{2z-1}=\frac{\left(z-\frac12+\frac12\right)^2+1}{2\left(z-\frac12\right)}=\frac18\cdot\frac{\left(1+2\left(z-\frac12\right)\right)^2+4}{z-\frac12}=$$$${}$$

$$=\frac1{8(z-\frac12)}\cdot\left(5+4\left(z-\frac12\right)+4\left(z-\frac12\right)^2\right)=\frac5{8\left(z-\frac12\right)}+\frac12+\frac12\left(z-\frac12\right)$$

Answer 2

Here we have the nice situation that the denominator already has (besides a multiplicative factor) the form $$z-\frac{1}{2}$$ and the approach of @DonAntonio is the most appropriate.

With respect to the general case you are addressing we can always take a geometric series.

• We obtain a power series expansion of $\frac{1}{z+a}$ at $z=z_0$ via \begin{align*} \frac{1}{z+a} &=\frac{1}{(z-z_0)-(a-z_0)}=-\frac{1}{a-z_0}\cdot\frac{1}{1-\frac{z-z_0}{a-z_0}}\\ &=-\sum_{n=0}^{\infty}\frac{1}{(a-z_0)^{n+1}}(z-z_0)^n \end{align*}

• The principal part of $\frac{1}{z+a}$ at $z=z_0$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{(z-z_0)-(a-z_0)}=\frac{1}{z-z_0}\cdot\frac{1}{1-\frac{a-z_0}{z-z_0}}\\ &=\frac{1}{z-z_0}\sum_{n=0}^{\infty}\frac{(a-z_0)^n}{(z-z_0)^n}\\ &=\sum_{n=1}^{\infty}\frac{(a-z_0)^{n-1}}{(z-z_0)^n}\ \end{align*}