# I^i has infinite values?

1. Mar 8, 2012

### Chantry

Can someone explain if this is true or if there's anything wrong with the following logic?

e^ix = cos(x) + isin(x)

Let x = pi/2 + 2npi
Then,
e^ix = i
Take both sides to the exponent i,
e^-x = i^i
e^-(pi/2 + 2npi) = i^i

But, e^(pi/2 + 2npi) has infinite different values.
I've been looking around on the internet and I can only find the typical e^(-pi/2) solution. Is it just that mathematicians have just decided to define it to be that value? Even wolfram alpha spits out only one value. Is it just as correct to say that i^i = e^-(5pi/2) for example?

Last edited: Mar 8, 2012
2. Mar 8, 2012

### morphism

First of all, it's not true that $(e^{ix})^i = e^{i^2x}$; exponentiation is more subtle in complex land. I'm sure if you search for "complex exponentiation" you'll find lots of hits (several on this forum alone, probably) that explain the various subtleties. Let me just point out that writing $z^i$ is already 'ambiguous'. This is because the former expression is defined to be $$e^{i \log z}$$ and the complex logarithm is multivalued. Thus the fact that you're getting infinitely many values for "i^i" is simply a manifestation of the fact that there are infinitely many possible values for log(z) at z=i.

3. Mar 9, 2012

### Chantry

To be clear, you're saying that you can't just simplify (a^i)^i to a^(i^2), BUT i^i does have infinite different values, correct?

This is starting to boggle my mind a little bit. I can't seem to wrap my head around it.

If you ctrl f and type in "Failure of power and logarithm identities" and look just above it, they do just that simplification of (a^i)^i to a^i*i. But then if you keep reading, below that proof they contradict themselves and say exactly what you're saying, which is that you can't simplify complex exponents in that way.

I don't know. Maybe I'll just revisit this when I've taken a course in complex analysis.

4. Mar 9, 2012

### Office_Shredder

Staff Emeritus
You only write down one distinct number, just various representations of it. Note e2npi=1 for all integers n

5. Mar 9, 2012

### phyzguy

This is the key point. Since e^(2*n*pi*i)=1, of course I can write the number this way. It's like saying that there are infinitely many results for 2^2, as follows:
4, 4*1, 4*(1^2), 4*(1^3), ...

6. Mar 9, 2012

### morphism

$e^{2n\pi}$ is most definitely not equal to 1 (unless n=0).

There are indeed infinitely many values that one can 'give' to the expression i^i, as mentioned in the OP.

7. Mar 10, 2012

### phyzguy

You're right. I withdraw my comment.