I want to find singular points and residue of the function $$f(z) = \frac{1}{\cos(z^{-2})} \;\;\; z\in \mathbb{C} $$
Can someone help me verify my solution? I found singularities to be points that are of the form $$a = \left(\frac{\pi}{2}+\mathbb{Z}\pi\right)^{-\frac12}$$ Since these are the points where cosine is zero. I gather that this is a rational function, so if f = h / g, this means that $$ h(a) \neq 0, \;\; g(a) = 0, \;\; g'(a) \neq 0$$ and the residue is $$ \operatorname{Res}(f,a) = \frac{h(a)}{g'(a)} = \frac{1}{2 \sin(\frac{1}{a^2})\frac{1}{a^3}} = \frac{1}{2\left(\frac{\pi}{2}+\mathbb{Z}\pi\right)^{\frac32}} $$ Is this correct?
The isolated singularities of $f$ are those numbers $\omega$ such that $\frac1{\omega^2}=\frac\pi2+n\pi$. And$$\operatorname{Res}_{z=\omega}f(z)=\frac1{2\sin^3\left(\frac1{\omega^2}\right)\frac1{\omega^3}}=\frac{(-1)^n\omega}{2\left(\frac\pi2+n\pi\right)}.$$