Does e^z - z^2 = 0 Have Infinite Solutions?

[Likemath2014]

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

 

[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 $e^z$.

 

[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 $n\in\mathbb{Z}$.

 

[WWGD]

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