# Complex exponential equation

1. Sep 24, 2014

### Likemath2014

How can we show that the following equation has infinitely many solutions
$$e^z-z^2=0$$.
Thanks

2. Sep 24, 2014

### WWGD

Work within a fixed branch of logz and write : $z^2=e^{2logz}$

Once you find a solution, you have infinitely-many, by periodicity of $e^z$.

3. Oct 7, 2014

### platetheduke

Continuing WWGD's post: you get $e^z=e^{2\log z}$ and thus the equation $e^{z-2log z}=1$. In other words you are looking at the solutions of each of the equations $z-2\log z=2\pi n$ for $n\in\mathbb{Z}$.

4. Oct 7, 2014

### WWGD

It seems that something like Lambert's W function may be helpful in finding solutions to platetheduke's equation.