Singularity at a given point

Question

If i have a function $f(z)=\frac{1}{sin(z^2)}$, in which order is the pole at $z=0$? I'm thinking that it's a simple pole, while it would have been a pole of order $2$ if $f(z)=\frac{1}{sin^2(z)}$. Is this correct?

Answer 1

No. The derivative of $\sin(z^2)$ is $2z \cos(z^2)$, which is zero at $z = 0$, but its second derivative, $2 \cos(z^2) - 4z^2 \sin(z^2)$ is $2$ at $z = 0$, so $z = 0$ is a zero of order $2$ of $\sin(z^2)$. From this, you can deduce the order of the pole of $\csc(z^2)$ at $z = 0$.

Answer 2

Since $1/\sin(z)$ has a simple pole at $z=0$ and is holomorphic around $z=0$, we have a representation: $$\frac{1}{\sin (z)}=\frac{a_{-1}}{z}+\sum_{n\ge 0} a_nz^n$$ Composing with $z^2$ yields: $$\frac{1}{\sin (z^2)}=\frac{a_{-1}}{z^2}+\sum_{n\ge 0} a_nz^{2n}$$ So in fact $1/\sin(z^2)$ has a pole of order $2$ at $z=0$.

Multiplying the series for $1/\sin(z)$ with itself yields the same result for $1/\sin^2(z)$