Determine the character of all singularities of function: $g(z)= \frac{1}{e^{z^{2}} -1}.$

Question

Determine the character of all singularities of function:

$$g(z)= \frac{1}{e^{z^{2}} -1}.$$

I would say $0$ is a pole, but I don't know how to indicate its order.

Answer 1

It is true that $z=0$ is a pole. To find its order use Taylor's expansion for $e^{z^2}$ at $z=0$ and you'll see that the leading term is $z^2$, hence the order of pole $0$ equals to $2$.

Answer 2

Actually one of the best way to find order of the pole of function kind of $\frac{1}{g(x)}$ - is to find first derivative, which is not equal to zero. In this case:

$g(0) = 0$

$g'(x) = 2e^{z^2}z$, so $g'(0) = 0$

$g''(x) = 2e^{z^2} + 2z^2e^{z^2}$, so $g''(0) = 2$

That's why the order is 2.

Actually, I also wanted to say, that not only z = 0 is singularity. Also $z^2 = 2\pi n i$.