I got this problem in my complex analysis class:
Find the Laurent series of $$f(z) = \frac{\sinh(z + 3i)}{z(z + 3i)^3}$$ to calculate the residue at $z=-3i$.
Is there an easy way to calculate the series and no leave it as a product of two series?
2026-03-26 00:56:34.1774486594
Laurent series for $f(z) = \frac{\sinh(z + 3i)}{z(z + 3i)^3}$ at $-3i$ (not leaving as a product of series)
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There's a removable singularity at $z=-3i$, so the residue there is zero.
If you really want to find the Laurent series, write $w=z+3i$. The function is now $$\frac1{w-3i}\frac{\sinh w}w =-\frac1{3i}\sum_{m=0}^\infty\frac{w^m}{(3i)^m} \sum_{n=0}^\infty\frac{w^{2n}}{(2n+1)!}.$$ I'd leave it like this ....