How many solutions has the equation $e^z=z$?

Question

How many solutions does the complex equation $$e^z=z$$ have ?

Can we somehow use that this equation implies $z=\ln(z)$ and if this is possible, how do we deal with the infinite many branches of the complex logarithm ?

Is $2$ the number of solutions ?

Also: Are there any simple ways to see that $e^z-z=0$ has infinitely many solutions? — Martin R, Jul 21 '19 at 08:32

score 1 · Accepted Answer · answered Jul 21 '19 at 08:21

The solution of the equation is given in terms of Lambert function $$z=-W_k(-1)$$ As you will see, there is no real solutions and, in the complex domain, there is an infinite number of solutions, each of them corresponding to the $k^{th}$ branch of Lambert function.

How many solutions has the equation $e^z=z$?

1 Answers1