Order of zeros and poles of $f(z) = \frac{e^{z^2} - 1}{z}$ at $z=0$ I just need clarification of a problem.
I am asked to find the order of zeros or poles for :
$$f(z) = \frac{e^{z^2} - 1}{z}$$
If I represent that as series, I get:
$$f(x) = z + \frac{z^3}{2!} + \frac{z^5}{3!} + \dots$$
Does this mean that there is a zero of order $1$ at $z= 0$ ?
 A: $z=0$ cannot be a zero of $f$ because $f$ is not defined there.
$z=0$ is a pole of order zero of $f$, also known as a removable singularity.
The extension of $f$ to an entire function has a zero of order one at $z=0$.
A: You don't need to go as far.
The definition of a pole is :

Suppose $U$ is an open subset of the complex plane $\mathbb C$, $p$ is an element of $U$ and $f : U-\{p\} → C$ is a function which is holomorphic over its domain. If there exists a holomorphic function $g : U → C$, such that $g(p)$ is nonzero, and a positive integer $n$, such that for all $z$ in $U-\{p\}:$
$$f(z)=\frac{g(z)}{(z-p)^n} $$
holds, then $p$ is called a pole of $f$. The smallest such $n$ is called the order of the pole. A pole of order $1$ is called a simple pole. A few authors allow the order of a pole to be zero, in which case a pole of order zero is either a regular point or a removable singularity. However, it is more usual to require the order of a pole to be positive.

That means that $z=0$ is a pole of order zero (removable singularity as mention in the definition (since you can see that in your case, $e^{z^2}-1 =0$ for $z=0$).
The extension of $f$ to an entire function has a zero of order $1$ at $z=0$ as you showed (note that $z=0$ can not be a root/zero of $f$ since $f$ is not defined at that point).
