# Showing that $z^3 e^z = 1$ has infinitely many solutions

On an old complex analysis prelim, I encountered the following problem. Show that the equation $z^3 e^z =1$ has infinitely many solutions. How many are real?

Well many sources in complex analysis utilize Picard's Big Theorem to show that $e^z -z =0$ has infinitely many solutions. Picard's Big Theorem states that if an analytic function $f$ has an essential singularity at a point $w$, then on any punctured neighborhood of $w$, $f(z)$ takes on every possible complex value, with at most one exception, infinitely often. Clearly, $f(z) = e^z -z$ has an essential singularity at $\infty$. Thus, $f$ takes on every value infinitely many times with at most one exception, at most one value can be achieved finitely often, and that value could possibly be $0$. But we recall that the exponential function is $2 \pi i$-periodic and so $f(z+ 2 \pi i)= f(z)- 2 \pi i$. So now $f$ takes on at least one value in $\{0,2 \pi i\}$ infinitely often. Applying Picard's Big Theorem to both these cases, implies that $e^z -z =0$ has infinitely many solutions.

Can Picard's Big Theorem also be used to show that $z^3 e^z =1$ has infinitely many solutions? We can rewrite this equation as $e^z - z^{-3} =0$ so that we could let $f(z)= e^z - z^{-3}$ and see that it has an essential singularity at $\infty$, correct? If so, then I think that maybe Picard's Big Theorem might be OK here, though I'm not really sure. Now the second part of the problem is to find how many solutions are real; there is a hint (hardly expected on a prelim) that suggests using the Argument Principle. I'm familiar with how to find the number of zeros in a particular quadrant/half plane of a complex polynomial of finite degree through use of the Argument Principle, but don't know how to apply it here. I'd appreciate any help I can get.

• How about consider $\frac{1}{z}$ transformation. If it has essential singularity, you might get accumulated singularity at 0 under $\frac{1}{z}$. So why not try integrating $\frac{f'}{f}$ around a contour about 0 of radius 1, under above transformation? If this diverges to -$infinity$, then you know it must have infinite number of poles. – user45765 Aug 15 '14 at 0:18

## 2 Answers

Can Picard's Big Theorem also be used to show that $z^3 e^z =1$ has infinitely many solutions?

Sure.

We can rewrite this equation as $e^z - z^{-3} =0$ so that we could let $f(z)= e^z - z^{-3}$ and see that it has an essential singularity at $\infty$, correct?

Yes, but it's more direct in the given form:

$$f(z) = z^3e^z$$

has only one zero, with multiplicity $3$. Hence $0$ is the exceptional value.

Note that the second part, concerning real solutions, is pretty trivial in means of elementary real calculus. For any negative $x$ we have $f(x)<0$ and on the positive ray $f$ is ascending.

• Does this mean that $z^3 e^z =1$ has only one real solution, i.e., the exceptional value $0$? – Libertron Aug 15 '14 at 0:23
• No, $x=0$ does not yield $x^3e^x=1$. However, since the root must be positive (because $e^x$ is always positive) and given the fact that the function is always increasing for $x>0$, there can be only one real root. Zero theorem then help you to find a reasonable interval where that is. – bartgol Aug 15 '14 at 1:33