# Are there any simple ways to see that $e^z-z=0$ has infinitely many solutions?

Joseph Bak and Donald Newman's complex analysis book (p.236) has a proof that the equation $e^z-z=0$ has infinitely many complex solutions:

I'm curious if there are any particularly elegant ways to see this, other than that given in the text.

-
Note that all the complex solutions of the equation come in the form $-W_k(-1)$, where $W_k(z)$ is the $k$-th branch of the Lambert function. Why the Lambert function has to be multibranched is a good question in itself. –  Guess who it is. Apr 28 '12 at 6:57

An elementary proof: Let $z = x +y i$ then $|e^z| = |z|$ precisely if $e^{2x} = x^2 + y^2$. If $x \geq 0$ then $e^{2x} - x^2 > 0$ so $y = (e^{2x} - x^2)^{1/2}$ is a positive solution of this equation. This means that for all $x \geq 0$ there is such a $y \geq 0$ such that $|e^z| = |z|$. The argument of $z$ is in $[0, \pi/2]$ since $x, y \geq 0$. The argument of $e^z$ is $y$. Since $y \to \infty$ when $x \to \infty$ there are infinitely many such $z$ for which both $|e^z|=|z|$ and $\arg(e^z) \equiv \arg (z) \pmod {2\pi}$. These $z$ are therefore roots of $e^z-z$.

-
Thanks, I really like this answer! –  Hana Bailey May 1 '12 at 5:43
@Cowbell Thanks. Not really in the spirit of complex analysis maybe, but it works. –  WimC May 1 '12 at 17:16

If you use the fairly deep result of Picard about essential singularities then you can prove this as follows: $f(z) = e^z-z$ has an essential singularity at infinity. Therefore $f$ attains all values infinitely many times with at most one exception (that is, at most a single value could be attained only finitely many times). This exception could still be $0$. However, $f$ also satisfies $f(z + 2\pi i) = f(z) - 2\pi i$. Now $f$ attains at least one value in $\{0, 2\pi i\}$ infinitely many times. In both cases it follows that $f$ must have infinitely many zeroes.

-

One way to see this is to realize that $f(z)=e^z$ has $\{z\in\mathbb{C}:0\leq\mathrm{Im}(z)<2\pi\}$ as a fundamental region and has period $2\pi i$. That fundamental region is mapped onto the plane (excluding $0$), as is every shift of the region by integer multiples of $2\pi i$. From there, it isn't difficult to show that there must be infinitely many $z\in\mathbb{C}$ for which $f(z)=z$.

It remains only to show (as pointed out below by Harald) that in each such shift of the region there is at least one solution--that is, at least one zero of the function $g(z)=e^z-z$. Harald's suggestion of applying the argument principle (see http://en.wikipedia.org/wiki/Argument_principle if needed) is a good one. Noting that the function is entire (so no poles), you really need only show that $\oint_C\frac{e^z}{e^z-z}dz$ is non-$0$ (where $C$ is the contour he suggests) for sufficiently large $M$.

-
It's a good proof, but I think the last part needs a bit of amplification. Basically, you need to see that there is a solution in every shifted copy of the fundamental strip. I think applying the argument principle to the boundary of a rectangle $-M\le\operatorname{Re}z\le M$, $2k\pi i\le\operatorname{Im}z\le2(k+1)\pi i$ for large $M$ and integer $k$ will do it. –  Harald Hanche-Olsen Apr 28 '12 at 7:58
Good point. I suppose that simply saying "it isn't difficult" isn't all that useful on its own. –  Cameron Buie Apr 28 '12 at 9:14
Thank you Cameron. –  Hana Bailey May 1 '12 at 5:44