# Need help

1. Feb 25, 2007

### racland

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

2. Feb 25, 2007

### neutrino

Have you at least made attempts to solve them?

3. Feb 25, 2007

### arildno

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

4. Feb 27, 2007

### 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...

5. Feb 28, 2007

### JonF

can you express the RHS in polar form?

6. Mar 1, 2007

### HallsofIvy

Staff Emeritus
Actually y= $\pi K$, since $sin(\pi)= 0$ also. Now, what values can cos y have? And then what is ex?

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).