Find the residue at all singular points

Question

I have the function $f(z) = \dfrac {\sin(z^2)}{z^2(z^2+1)}$ and I need to identify the singular points and find the residue at each point. I've identified $z=0$ as a removable singularity, and $z=i$ and $z=-i$ as simple poles. For the simple poles, I calculated the residue to be zero for each. However, I'm not sure how to find the residue at $z=0$. Should I formulate a power series centered at $z=0$ and just see what the coefficient of $\frac {1}{z}$ is?

Answer 1

Note that for $f(z)=\frac{\sin(z^2)}{z^2}$ we can write

$$f(z)=\sum_{n=0}^\infty \frac{(-1)^n(z^2)^{2n+1}}{z^2(2n+1)!}=\sum_{n=0}^\infty \frac{(-1)^nz^{4n}}{(2n+1)!} \tag 1$$

Evidently, we see that $f(z)$ is analytic and its residue is, therefore, $0$.

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Alternatively, we see that $\lim_{z\to 0}\frac{\sin(z^2)}{z^2}=1$ and hence there is no singularity (other than a removable one) at $z=0$.

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For $g(z)=\frac{f(z)}{z^2+1}$, the reside at $z=\pm i$ can be computed as the limit

$$\lim_{z\to \pm i}\frac{f(z)}{z\pm i}=\frac{f(\pm i)}{\pm2 i }=\frac{\sin(1)}{\pm 2i}$$

And the residue at $z=0$ is $0$ since the singularity is removable.