I tried for a while to classifiy the singularities of $\frac{1}{z}-\frac{1}{\sin z}$ at the origin, but I am stucked. Is there someone who is able to help me at this point?
2026-03-25 23:16:28.1774480588
$\frac{1}{z}-\frac{1}{\sin z}$ at the origin - Classify singularities
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Hint: $\sin(z)=z\bigl(1+z^2g(z)\bigr)$ where $g$ is analytic with $g(0)\ne0$.
Edit: So $$\frac1z-\frac1{\sin z}=\frac{\sin z-z}{z\sin z}=\frac{z^3g(z)}{z^2\bigl(1+z^2g(z)\bigr)}.$$ Can you take it from there?