Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

enter image description here

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

share|cite|improve this question
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. – J. M. Apr 28 '12 at 6:57
up vote 6 down vote accepted

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 $e^z-z$.

share|cite|improve this answer
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.

share|cite|improve this answer

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 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 $M$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.