# Homework Help: Very simple complex powers problem

1. Sep 24, 2012

### Aziza

Write the following in x+iy form: (-1)^i

my solution:
(-1)i = (ei(π+2πk))i = e-(π+2πk)

However, my book just states the answer as e-(π) .....but these are not the same....

2. Sep 24, 2012

### Muphrid

For integer $k$, sure they are. Any full turn around the unit circle doesn't change the value.

3. Sep 24, 2012

### Aziza

but ei2πk and e2πk are different...the first one equals 1 but the second does not...

i mean look just plug in k=1 and k=2...you get e^-3 and e^-5, which are not the same

4. Sep 24, 2012

### Muphrid

Oh, okay, I see what's going on. Yeah, in this case, you may be more correct than the book is. Someone with more recent experience than I in complex analysis may know something more specific to this, though.

5. Sep 24, 2012

### Aziza

oh alright. idk my book is doing this same mistake for each and every related problem. so i was thinking that i must be doing something wrong. but maybe they are just giving the answer when k=0...although they never specify this....

6. Sep 24, 2012

### Curious3141

Your answer is more complete than the book's. Exponentiation to a complex power can give a multivalued result. The book just gave the principal value, which, if you're asked for a single answer, is the usual value we take.

7. Sep 24, 2012

### SammyS

Staff Emeritus
(Those π's are hard to read !)

Sure enough.

Give WolframAlpha, $\displaystyle \left(e^{\pi} \right)^{1/i}$ or $\displaystyle \left(e^{-3\pi} \right)^{1/i}\,,$ and it gives -1.

Of course, $\displaystyle e^{-3\pi}\approx 0.0000806995175703\,,$ $\displaystyle e^{-\pi}\approx 0.04321391826377226\,,$ $\displaystyle e^{\pi}\approx 23.140692632779263\,,$

So it looks like $\displaystyle (-1)^i=e^{(\text{odd integer multiple of }\pi)}\ .$