Laurent series to converge in $0<|z-1|<R$

Question

Question:

Determine the largest number $R$ so that the Laurent series of $$f(z)=\frac2{z^2-1} + \frac3{2z-i}$$ about $z=1$ converges for $0<|z-1|<R$.

Attempt:

I really don't understand this question. The singularities are $\pm 1, i/2$. The point $1$ doesn't lie in the region given. If $R$ was such that $-1$ would be a boundary point on the circle, $R=2$. And if $R$ was such that $i/2$ would be a boundary point on the circle, $R=\sqrt5/2$. Both can't be a boundary point but the largest $R$ is $2$ so would that simply be the answer?

I don't see how this would make the series converge either. I really don't understand this question!

Answer 1

Two facts about Laurent series are important here:

1. If $h$ is holomorphic in an annulus $r < \lvert z-a\rvert < R$, then there is a Laurent series $$\sum_{n=-\infty}^\infty c_n (z-a)^n$$ converging to $h$ in that annulus, and
2. the sum function of a Laurent series is holomorphic in its annulus of convergence, i.e. if $$s(z) := \sum_{n=-\infty}^\infty c_n (z-a)^n$$ converges for $r < \lvert z-a\rvert < R$, then $s$ is holomorphic in the annulus $r < \lvert z-a\rvert < R$.

In our case, where the inner radius of the annulus is $0$, that means the outer radius of the annulus is the distance to the closest singularity to the centre of the Laurent series.

The two singularities of $f$ other than $1$ are the poles at $-1$ and $\frac{i}{2}$. Since $\frac{i}{2}$ is closer to $1$ than $-1$ is, we have

$$R = \biggl\lvert \frac{i}{2} - 1\biggr\rvert = \frac{\sqrt{5}}{2},$$

as you found in the comments.