I know there is a singularity of $z=1$ but I am a bit confused on how to find the residue at that point since if we have that $f(z)=\frac{g(z)}{h(z)}$ with $g(z)=1$ and $h(z)=(1-z)^3$ then $g(z)$ has a zero of order 0 (no zero) and $h(z)$ has a zero of order 3 since its first and second derivatives vanish at $z=1$ but the third derivative does not vanish at $z=1$. That is all I have, any help would be appreciated!
2026-03-28 10:24:19.1774693459
Residue of $\frac{1}{(1-z)^3}$ at $z=1$
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