The complex exponential function

[James889]

Hi,

I need to solve the equation
e^z = -3

The problems arises when i set z to a+bi
e^a(cos(b) + isin(b),~b = 0

Then I am left with e^a = -3

However you're not allowed to take the log of a negative number.

Also i know that cos(\pi) + isin(\pi) = -1

Obviously 3e^{(\pi*i)} is a solution, but it's not a solution for z.

Any ideas?

 

[HallsofIvy]

No, if you set it in that form, e^z= e^{a+ bi}= e^ae^{bi}= e^a(cos(\theta)+ i sin(\theta))= -3 but now e^a must be positive. We must have e^acos(\theta)= -3 and e^a sin(\theta)= 0. Since e^a is never 0, we must have sin(\theta)= 0 so that \theta= 0 or \pi. If \theta= 0, cos(\theta)= 1 so e^a= -3 which, as I said, is impossible. Therefore, \theta= \pi. Now continue.