Finding the Solutions to exp(z) = 1 with Complex Numbers

[fraggy]

Homework Statement

solving the equation e^z = 1 , where z is complex

The Attempt at a Solution

This is my attempt at the solution

e^z = e^a+ib

and then i can write it as e^a(cos(b)+isin(b))= 1
e^a 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 e^a 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$\pi$

 

[Dick]

Homework Statement

solving the equation e^z = 1 , where z is complex

The Attempt at a Solution

This is my attempt at the solution

e^z = e^a+ib

and then i can write it as e^a(cos(b)+isin(b))= 1
e^a 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 e^a 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$\pi$

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?

 

[fraggy]

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

 

[SteamKing]

misstake is a mistake