Question:
Determine the largest number $R$ such that the Laurent series of
$$f(z)= \dfrac{2}{z^2-1} + \dfrac{3}{2z-i}$$
about $z=1$ converges for $0<|z-1|<R$?
Attempt:
The radius of convergence of the Laurent series about one of its poles is equal to the distance to the nearest neighboring pole.
In the case of interest, $f(z)= \dfrac{2}{z^2-1} + \dfrac{3}{2z-i}$, and there are three poles; $z=1$, $z=\dfrac{i}{2}$, and $z=-1$. The distances between the original pole at $z=1$ and the other poles are $|1-(-1)|=2$, and $|1-\biggr(\dfrac{i}{2}\biggl)|=\sqrt{\dfrac{5}{4}}<2$.
Therefore, the Laurent series of $f(z)$ around $z=1$ has a radius of convergence $R=\sqrt{\dfrac{5}{4}}<2$. Is this correct, or, since the question asked for the largest number $R$, would R not be $=2$?
The function has poles at $\;z=\pm1\,,\,\,\frac i2\;$ , so the maximal convergence radius of convergence around $\;z=1\;$ cannot be greater than
$$\left|1-\frac12i\right|=\sqrt{1+\frac14}=\frac{\sqrt5}2$$
Together with my comment I think this explains why $\;R=2\;$ cannot be.