Find the residue, state the nature of the singularity, find the constant term in $1/sin(ze^z)$ at $z=0$
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Find the residue, state the nature of the singularity, find the constant term in series $1/sin(ze^z)$ at $z=0$ . We can rewrite the function $frac{1}{sin(ze^z)}$ as $frac{ze^z}{sin(ze^z)}cdotfrac{1}{ze^z}$ . What I did next I think might not be true and that's why I'm writing this post. Since $limlimits_{wrightarrow0}frac{w}{sin w}=1$ and $limlimits_{zrightarrow 0}ze^z=0$ and $[{dover dz}e^z]_{z=0}=1$ basically I said that at zero we can just look at $frac{1}{ze^z}={1over z}+sumlimits_{n=1}^inftyfrac{(-1)^n}{n!}z^{n-1}$ as the Laurent series of our original function. But I feel like this is far from rigorous... From this it results that the residue at zero is $1$ , which is true for the original function; the constant term is $-1$ , also true for $frac{1}{sin(ze^z)}$ an