# Solving exp(z) = 1

1. Mar 12, 2014

### fraggy

1. The problem statement, all variables and given/known data

solving the equation ez = 1 , where z is complex

3. The attempt at a solution

This is my attempt at the solution

ez = ea+ib

and then i can write it as ea(cos(b)+isin(b))= 1
ea has to be larger than 0 and therefor i*sin(b) = 0 → b can either be n2$\pi$ or $\pi$ + n2$\pi$
and cos(b) has to be positive because ea is positive, and the only posible b is n2$\pi$. And because a is the real part a has to be equal to 0. I really can't see where I'm missing something because in the solution z = in2$\pi$

2. Mar 12, 2014

### Dick

All fine. So b has to be 2*pi*n and a has to be 0. So z=a+i*b=i*2*pi*n. What's the problem?

3. Mar 12, 2014

### fraggy

Ok i got it now thx. I forgot to factor in the i with b. Wow really simple misstake

4. Mar 13, 2014

### SteamKing

Staff Emeritus
misstake is a mistake