TITLE: Are there any simple ways to see that $e^z-z=0$ has infinitely many solutions?
QUESTION [8 upvotes]: 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.

REPLY [9 votes]: 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 $e^z-z$.