# Find solutions of e^z = 1

Complex Analysis:
Find all solutions of:
(a) e^z = 1
(b) e^z = 1 + i

Have you at least made attempts to solve them?

arildno
Homework Helper
Gold Member
Dearly Missed
How fascinating!
We are all desirous to see what you have done so far!

So far I got:
e^z = 1
I know e^z = e^x (cos y + i Sin y)
Then,
e^x cos y = 1
e^x sin y = 0
I know that for sin y = 0, y = 2 pi * K, where k is an integer.
After this step I'm stuck!

The second problem: e^z = 1 + i
e^x cos y = 1
e^x sin y = 1
I order for sin y = 1, y = pi/2. Now what...

can you express the RHS in polar form?

HallsofIvy
Homework Helper
So far I got:
e^z = 1
I know e^z = e^x (cos y + i Sin y)
Then,
e^x cos y = 1
e^x sin y = 0
I know that for sin y = 0, y = 2 pi * K, where k is an integer.
After this step I'm stuck!
Actually y= $\pi K$, since $sin(\pi)= 0$ also. Now, what values can cos y have? And then what is ex?

The second problem: e^z = 1 + i
e^x cos y = 1
e^x sin y = 1
I order for sin y = 1, y = pi/2. Now what...
No, exsin y= 1 does NOT tell you that sin y= 1! You might try this: square each equation and add. That will get rid of y so you can find ex (you don't really need to find x itself).