<s>[INST] Classifying singular points of $\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}$ 
I am trying to classify the singular points of the function $$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}.$$

My attempt:
$$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}=\frac{\sin(z^2)}{z^2\left(z-\frac{\pi}{4}\right)}.$$
Hence the singular points are $z=0,\frac{\pi}{4}$. For classification:
$z=\frac{\pi}{4}$ is a simple pole as $\left(z-\frac{\pi}{4}\right)$ is a simple zero.
For $z=0$, we can see that $z^2\left(z-\frac{\pi}{4}\right)$ is a zero of order $2$. Also,$\ \sin(z^2)$ is a zero of order $2$ for $z=0$. Hence, $z=0$ is a removable singularity. 
I am unsure about the $z=0$ case, particularly if $\sin(z^2)$ is a zero of order $2$ for $z=0$.
 [/INST] Note that $$\sin(z^2)=z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}-\cdots$$ so $$f(z)=\frac{z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}-\cdots}{z^2\left(z-\frac{\pi}{4}\right)}=\frac{1-\frac{z^4}{3!}+\frac{z^8}{5!}-\cdots}{z-\frac\pi4}$$ so $z=0$ is a removable singularity.
</s>