Show that $f(z)=ze^{\frac{1}{z}}e^{\frac{-1}{z^2}}$has an essential singularity at $z=0$.

This one should be straightforward, as I should be able to tackle it by use of expanding the power series for $e^{1/z}$ and $e^{-1/z^2}$ and multiplying them out (or alternatively expanding $ze^{\frac{z-1}{z^2}}$).

But a singularity is essential if and only if the Laurent expansion around that point has an infinite number of terms with negative exponents, and the challenge here, informally, is to see whether the product of the series doesn't in fact end up "adding out to zero" for each negative-powered term after some particular $a_nz^{-n}$ in the Laurent expansion resulting from the product of the individual Laurent expansions for each term in the function.

My solution would be to attempt to write the coefficients of the negative powers as a general formula that I would end up pointing out must be nonzero for an infinite number of negative powers, via summing all the coefficients for a general $z^{-n}$ term in the product by use of the binomial expansion, but my efforts haven't amounted to much as far as this explicit calculation goes.

I have, then, thought about alternative approaches:

(1) Attempt to show that $f(1/z)=\frac{e^{(z-z^2)}}{z}$ has infinitely many positive terms when expanded around $z=\infty$. This runs into the same problem as before, except without fractions.

(2) Attempt to show, given $\epsilon >0$, that $f(z)$ on $D(0,\epsilon)$ is dense in $\mathbb{C}$. I haven't moved an inch on this one.

(3) Showing that the product of two functions with an essential singularity at the same point is a function with an essential singularity at that same point. I aborted this path after realizing $e^{1/z}*e^{-1/z}$ would be a counterexample.

Is there something I'm not seeing here? Any help is appreciated!

  • $\begingroup$ On the subject of your second alternative approach, I think it would suffice to just show there is a sequence $z_n\to0$ such that $f$ takes the same value on each $z_n$. $\endgroup$ – Trevor J Richards Mar 6 '14 at 0:41
  • $\begingroup$ I am not familiar with that theorem. Do you have the conditions that show such an equivalence? Certainly it would not hold in general e.g. $f(z)=0$ has no essential singularity at $z=0$ but satisfies your suggestion for eg $z_n=1/n+0i$. $\endgroup$ – Darrin Mar 6 '14 at 0:52
  • 1
    $\begingroup$ Darrin, The theorem is as follows: If $f$ is analytic, and the set $\{z:f(z)=c\}$ has an accumulation point in the domain of $f$, then $f$ is constant, $f(z)=c$. Therefore since your $f$ is obviously not constant, if you can find the $z_n$ as mentioned before, it would show that neither $f$ nor $1/f$ is analytic at $0$. Thus $0$ must be an essential singularity (since it is a singularity and not a pole). $\endgroup$ – Trevor J Richards Mar 6 '14 at 1:09

If the singularity is not essential, then there is a positive integer $n$ such that $\lim_{z\to 0} z^n f(z)=0$. Plugging $z=i/k$ into $|z^n f(z)|$ yields $$ |z^n f(z)| = \frac{e^{k^2}}{k^{n+1}} $$ Which tends to $\infty$ (e.g., raise to power ${1/k}$ to get $e^k$ over something with limit $1$).

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