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### Complex Powers

```Date: 04/10/2002 at 16:43:02
From: Brian England
Subject: Complex exponents

I recently came across an interesting result and was wondering if
you could help.

Given e^(2*pi*i/2*pi*i) = e^(1) = e

but e^(2*pi*i/2*pi*i) = (e^(2*pi*i))^(1/2*pi*i)
and we know e^(2*pi*i) = 1
so 1^(1/2*pi*i) has to be equal to e.

I am having trouble proving this last step. I have rewritten the
equation many times, but keep getting stuck.

Thanks,
Brian England
```

```
Date: 04/10/2002 at 22:48:28
From: Doctor Peterson
Subject: Re: Complex exponents

Hi, Brian.

Part of the problem here is that complex powers are multi-valued;
there is not just one value of i^i, for example (though we often give
one answer, it is really just the principal value, much as we give one
square root of 2 when there are really two of them).

For example, since

1 = e^(2 k pi i)  for any integer k

we can say that

1^i = e^(2 k pi i i) = e^(-2 k pi)

Of these infinitely many values, the principal one is for k = 0,
namely 1 (of course). What's interesting here is that, whereas we say
that 1 to any power is 1, for complex exponents that is just one of
the possible values.

Your case is really just this raised to the -1/(2 pi) power:

1^(1/(2 pi i)) = 1^(-i/(2 pi))

= (1^i)^(-1/(2 pi))

= e^(-2k pi * -1/(2 pi))

= e^k

FOR ANY INTEGER k.

In other words, although, as you saw, your expression gives e, it also
gives the expected 1 (for k = 0) and all other powers of e.

I don't recall ever applying this idea to powers of 1 before. Maybe I
just couldn't bear the weirdness of it all!

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers

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