Does $\frac {z^5}{\sin z^2-z^2}$ have a non-isolated singularity at $0$? If so, is it not meaningful to discuss its residue at $0$?
2026-03-28 21:06:23.1774731983
Does $\frac {z^5}{\sin z^2-z^2}$ have a non-isolated singularity at $0$?
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As you have holomorphic/ holomorphic it shouldn't be non-isolated (you need a accumulation point of zeroes in the denominator for a non isolated singularity) and taking $$\sin(z^2)=z^2 -\frac{z^6}{3!}\pm \dots$$ gives you $$\frac{z^5}{z^2-z^2 - \frac{z^6}{3!}\pm \dots}= \frac{1}{-\frac{z}{3!}\pm\dots}$$ which has a pole of order $1$ at zero.
If the singularity is not isolated there is no sense in calculating the residue (which I guess won't be possible at all) because all the nice theorems won't work. In the definition of my complex analysis course we need the poles to be discrete (having no accumulation point) else we won't say it is a residue anyway.