Krdytkyu

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 $frac{1}{sin(ze^z)}$ and $z_0=0$ is a pole of degree $1$ also true.

The function $g(z)=ze^z$ is analytic and $g'(z)=e^z+ze^z$ is different from $0$ in a neighborhood of $0$, so it is invertible on that neighborhood, with analytic inverse.

Thus the substitution $w=ze^z$ is possible in a limit and
$$
lim_{zto0}frac{z}{sin(ze^{z})}=
lim_{zto0}frac{ze^z}{sin(ze^{z})}e^{-z}=
lim_{wto0}frac{w}{sin w}e^{-g^{-1}(w)}=1
$$
Therefore $f(z)=1/!sin(ze^{z})$ has a pole of order $1$ at $0$ and the residue is $1$.

The derivative of $zf(z)$ (for $zne0$) is
$$
frac{sin(ze^z)-z(e^z+ze^z)cos(ze^z)}{sin^2(ze^z)}=
frac{sin w-wcos w-g^{-1}(w)wcos w}{sin^2w}
$$
Note that $g^{-1}(w)=w+o(w)$, because its derivative at $0$ is $1$, so we have
$$
lim_{zto0}frac{sin(ze^z)-z(e^z+ze^z)cos(ze^z)}{sin^2(ze^z)}=
lim_{wto0}frac{w-w-w^2+o(w^2)}{w^2+o(w^2)}=-1
$$
Thus
$$
zf(z)=1-z+o(z^2)
$$
and finally
$$
f(z)=frac{1}{z}-1+o(z)
$$

With the help of a computation package, I found that, at $z=0$,
$$
sin(zexp(z)) = z + z^2 + frac13z^3 - frac13z^4 - frac7{10}z^5 + cdots,,
$$
and its reciprocal is
$$
frac1{sin(zexp(z))} =
z^{-1} - 1 + frac23z + frac{13}{90}z^3 + frac7{90}z^4 + frac{37}{378}z^5 +cdots,.
$$
You could certainly determine the first few terms of these by mindless hand-computation.

I think, though, that seeing that the residue is $1$ is just a matter of direct examination without any kind of computation at all.

You can find the first few terms of the series, step-by-step

$$ ze^z = zleft(1 + z + frac{z^2}{2} + dots right) = z + z^2 + frac{z^3}{2} + dots $$

begin{align} sin(ze^z) &= sinleft(z + z^2 + frac{z^3}{2} + dots right) \
&= left(z + z^2 + frac{z^3}{2} dots right) - frac{1}{3!}left(z + z^2 + frac{z^3}{2} +dots right)^3 + dots \
&= z + z^2 + frac{z^3}{2} - frac{z^3}{6} + dots \
&= z + z^2 + frac{z^3}{3} + dots end{align}

begin{align}
frac{1}{sin(ze^z)} &= frac{1}{z+z^2+frac{z^3}{3}+dots} \
&= frac{1}{z} frac{1}{1 + z + frac{z^2}{3} + dots} \
&= frac{1}{z} left[1 - left(z + frac{z^2}{3} + dotsright) + left(z + frac{z^2}{3} + dotsright)^2 - dots right] \
&= frac{1}{z}left[1 - z - frac{z^2}{3} + z^2 + dotsright] \
&= frac{1}{z}left[1 - z + frac{2z^2}{3} + dots right]
end{align}

so the residue is $1$ and the constant term is $-1$