What are the solutions for e^z = 1 and e^z = 1 + i?

[racland]

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

 

[neutrino]

Have you at least made attempts to solve them?

 

[arildno]

How fascinating!
We are all desirous to see what you have done so far!

 

[racland]

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...

 

[JonF]

can you express the RHS in polar form?

 

[HallsofIvy]

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 e^x?

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, e^xsin 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 e^x (you don't really need to find x itself).