# Help me solve this equation involving exponentials!

AxiomOfChoice
I'm trying to solve the following equation for $$z\in \mathbb C \setminus \{ 0 \}$$, $$w\in \mathbb C$$:

$$e^{1/z} + \frac{1}{1-e^{1/z}} = w.$$

How in the world should I go about doing that?

Homework Helper
Let u= $e^{1/z}$. Then your equation becomes
$$u+ \frac{1}{1- u}= w$$
Multiply both sides by 1- u to get u(1- u)+ 1= w(1- u) or $u- u^2+ 1= w- uw$ which equivalent to the quadratic equation $u^2- (1+w)u+ w-1= 0$. Use the quadratic formula to solve that, then solve $e^{1/z}= u[/math] by taking the logarithm of both sides. AxiomOfChoice Let u= [itex]e^{1/z}$. Then your equation becomes
$$u+ \frac{1}{1- u}= w$$
Multiply both sides by 1- u to get u(1- u)+ 1= w(1- u) or $u- u^2+ 1= w- uw$ which equivalent to the quadratic equation $u^2- (1+w)u+ w-1= 0$. Use the quadratic formula to solve that, then solve [itex]e^{1/z}= u[/math] by taking the logarithm of both sides.

Great. Thanks. This was my original approach, but for some reason I wasn't sure if I could make that change of variables and apply the quadratic formula like you did. But you've confirmed my intuition, so I'm going with it!

ande