Help me solve this equation involving exponentials!

  • #1

Main Question or Discussion Point

I'm trying to solve the following equation for [tex]z\in \mathbb C \setminus \{ 0 \}[/tex], [tex]w\in \mathbb C[/tex]:

[tex]
e^{1/z} + \frac{1}{1-e^{1/z}} = w.
[/tex]

How in the world should I go about doing that?
 

Answers and Replies

  • #2
HallsofIvy
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Let u= [itex]e^{1/z}[/itex]. Then your equation becomes
[tex]u+ \frac{1}{1- u}= w[/tex]
Multiply both sides by 1- u to get u(1- u)+ 1= w(1- u) or [itex]u- u^2+ 1= w- uw[/itex] which equivalent to the quadratic equation [itex]u^2- (1+w)u+ w-1= 0[/itex]. Use the quadratic formula to solve that, then solve [itex]e^{1/z}= u[/math] by taking the logarithm of both sides.
 
  • #3
Let u= [itex]e^{1/z}[/itex]. Then your equation becomes
[tex]u+ \frac{1}{1- u}= w[/tex]
Multiply both sides by 1- u to get u(1- u)+ 1= w(1- u) or [itex]u- u^2+ 1= w- uw[/itex] which equivalent to the quadratic equation [itex]u^2- (1+w)u+ w-1= 0[/itex]. 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!
 
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