1. ## e^z=4?

How do I solve e^z=4 and e^z=1, where z is complex number?

2. Originally Posted by hohoho00
How do I solve e^z=4 and e^z=1, where z is complex number?
write $e^z$ as $e^{x + iy} = e^x \cdot e^{iy}$

now note that you can write $4 = 4e^{i(2k \pi)}$ and $1 = e^{i(2k \pi)}$ for some integer $k$.

so your first equation is: $e^x \cdot e^{iy} = 4e^{2k \pi i}$

and your second equation is: $e^x \cdot e^{iy} = e^{2k \pi i}$

can you finish?

3. So, here's what I'm having trouble with:

so for e^x+1=0, I knew that if I let e^x = 1 and e^(iy) or (cosy+ isiny) = 2(pi)n, I can have solutions. But how do I know that is the only solutions? That is, how do I know that there aren't some e^x(cosy) =1 that aren't listed by above method? For example, some e^x bigger than 1 and some cosy smaller than 1 and somehow make up the number 1? ... Oh wait... I think I just answered my own question.