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.