Classification of the singularities of $\frac{1}{z^2+1}$ at $a=1$

Question

I have to classify the singularities of $\frac{1}{z^2+1}$ at $a=1$ and compute the radius of convergence of his power serie at the point $a$.

Is it possible to proceed without using the Laurent serie directly?

Thanks!

Answer 1

Factoring, $$ \frac{1}{z^2 + 1} = \frac{1}{(z-i)(z+i)} $$ By definition, $f$ has a pole of order $m$ at $z_0$ if there exists a function $g$ that is holomorphic in a neighbourhood of $z_0$ with $g(z_0) \neq 0$ such that $$ (z-z_0)^m f(z) = g(z) $$ so at $z_0 = -i$, we can choose $m = 1$, $g(z) = \frac{1}{z-i}$. This shows that $f$ has a simple pole at $z_0 = -i$, and likewise for $z_0 = +i$.

If you let the centre of the power series be $a = 1$, since the function $\frac{1}{z^2 + 1}$ is holomorphic in the disk centred at $a$ with radius $\sqrt{2}$, the series should have radius of convergence of $\sqrt{2}$, by Laurent's theorem.