Classifying singular points of $\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}$

Question

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 $z=0$.

Answer 1

It is correct.

Also you can notice that

$$\lim_{z\to 0} \frac{\sin(z^2)}{z^2\left(z-\frac{\pi}{4}\right)}= \lim_{z\to 0} \frac{\sin(z^2)}{z^2} \cdot \lim_{z\to 0}\frac1{z-\frac\pi4} = 1 \cdot \frac1{-\frac\pi4} = -\frac4\pi$$

so $z = 0$ is a removable singularity.

Answer 2

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.