I'm trying to solve an exercise about the classification of singolarity of a function.
$f(z)=\frac{e^{iz+1}-1}{(z^2+1)^2}$
I have found two poles. A first order pole in $i$ and a second order pole in $-i$. There should also be an essential singularity. How can I calculate it? Thank you so much.
The function has a essential singularity at infinity, you can see this by noting that the poles can be eliminated by subtraction with rational functions that is:
$$\phi(z) = f(z) - q(z)$$
have no finite singularities, consequently it can be Taylor expanded with infinite convergence radius. That is:
$$\phi(z) = \sum c_kz^k$$
Now either the sum is finite (in which case we have a pole at infinity or it's constant) or is infinite (in which case we have a essential singularity at infinity).
Suppose it were finite then we would have that $\phi(z)$ were a polynomial and that
$$e^{iz+1} = (z+1)^2 f(z) + 1 = (z+1)^2 (\phi(z)+q(z)) + 1$$
that is $e^z$ would be a rational function and since $e^z$ have no poles it would be a polynomial which we know it isn't. So we can conclude the Taylor expansion to be infinite and we have a essential singularity at infinity.