How to find the largest $R$ such that the Laurent series of $f(z)=\frac{2}{(z^-1)}+\frac{3}{2z-i}$ about $z=1$ converges for $0<|z-1|<R$?

Question

How to find the largest $R$ such that the Laurent series of $$f(z)=\frac{2}{(z^2-1)}+\frac{3}{2z-i}$$ about $z=1$ converges for $0<|z-1|<R$?

What I have done so far: $$\frac{2}{(z^2-1)}=\frac{2}{(z+1)(z-1)}=\frac{1}{(z-1)}-\frac{1}{(z+1)}$$ so the Laurent series for the $\frac{1}{(z-1)}$ part is just $\frac{1}{(z-1)}$ valid for all $z\neq 0$. Then the Laurent series for $$-\frac{1}{(z+1)}$$ is given by $$-\sum_{n=0}^{\infty} \frac{2^{n+1}}{(z-1)^{n+1}}$$ valid for $2<|z-1|$.

Now I am stuck, am I even on the right lines?

Answer 1

Being given a fonction f(z) with poles, if you want to have a clear and rapid answer of how are the annulli in which you have a specific Laurent expansion centered in a certain $z_0$, here is the rule:

Let $z_1,z_2,\cdots z_n$ be the poles of $f(z)$ ordered by increasing distance $d_1,d_2,\cdots d_n$ from $z_0$, then the radii of these annular regions are these $d_k$s, with possible empty annuli.

In the case at hand, $z_1=i/2, z_2=-i/2, z_3=-1$. Thus the radius you are looking for is $d_1=|z_0-z_1|=\sqrt{5}/2$.

Here is a figure that explains it, in this case, and in the case where the Laurent development is around point $-1-i$ (annuli delimited by dotted circles)

Answer 2

In general if $f$ is holomorphic on $\Omega \backslash \{z_0\}$, the radius for the Laurent's series is $R=dist(z_0,\mathbb{C}\backslash \Omega)$.

Here, you have singularities at $z= 1$, $z=-1$ and $z=\frac{i}{2}$. Hence by Laurent's theorem the $R$ you're searching for is $$R = \min\{|1-(-1)|,|1-\frac{i}{2}|\}=\min\{2,\frac{\sqrt{5}}{2}\}=\frac{\sqrt{5}}{2}.$$