A while back, I was studying for quals and I came across the problem:

Express [imath]i^i[/imath] using only real numbers.

The given solution was to take [imath]i = e^{i\pi/2}[/imath], so

[math]i^i = (e^{i\pi/2})^i = e^{-\pi/2}[/math]

But wait a minute, [imath]e^{i5\pi/2} = i[/imath] as well. Couldn't we just as easily use that to get [imath]i^i = e^{-5\pi/2}[/imath]?

But then you could apply similar "logic" to any expression with a complex exponent. Take the beloved identity [imath]e^{2\pi i} = 1[/imath]. We can use it to write

[math]e = e^{2\pi i}*e = e^{2\pi i + 1}[/math]

and then reapply this to itself to get

[math]e^{2\pi i} = (e^{2\pi i + 1})^{2\pi i} = e^{-4\pi ^{2}}e^{2\pi i}[/math]

This looks like nonsense, but I don't see where there's any mistake along the way. What's the deal? Is any expression with a complex exponent actually multivalued, and our choice just a matter of convention? Or is there some special rule I'm missing? If nothing else, this should give us another way to "prove" 1 = 0, or 2 + 2 = 5.

## i^i

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- skeptical scientist
- closed-minded spiritualist
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### Re: i^i

How do you define x

^{y}? It's common to define e^{y}by means of a power series, ln to be the inverse of the exponential function, and then define x^{y}as e^{y ln x}. This approach is great as long as x is real, but ln is not single-valued on complex numbers, so doesn't really work as a definition when your base is complex. Similarly, we could define x^{n}as x*x*x*...*x (n times) for positive integers n, 1 if n=0, or 1/x^{-n}if n is a negative integer, and this works for all nonzero x, but only if the exponent is an integer. So there's not really a good consistent way to define x^{y}when both x and y are allowed to be complex, which is why you had trouble with i^{i}.I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

"With math, all things are possible." —Rebecca Watson

### Re: i^i

Vicious Chicken wrote:A while back, I was studying for quals and I came across the problem:

Express [imath]i^i[/imath] using only real numbers.

The given solution was to take [imath]i = e^{i\pi/2}[/imath], so

[math]i^i = (e^{i\pi/2})^i = e^{-\pi/2}[/math]

But wait a minute, [imath]e^{i5\pi/2} = i[/imath] as well. Couldn't we just as easily use that to get [imath]i^i = e^{-5\pi/2}[/imath]?

But then you could apply similar "logic" to any expression with a complex exponent. Take the beloved identity [imath]e^{2\pi i} = 1[/imath]. We can use it to write

[math]e = e^{2\pi i}*e = e^{2\pi i + 1}[/math]

and then reapply this to itself to get

[math]e^{2\pi i} = (e^{2\pi i + 1})^{2\pi i} = e^{-4\pi ^{2}}e^{2\pi i}[/math]

This looks like nonsense, but I don't see where there's any mistake along the way. What's the deal? Is any expression with a complex exponent actually multivalued, and our choice just a matter of convention? Or is there some special rule I'm missing? If nothing else, this should give us another way to "prove" 1 = 0, or 2 + 2 = 5.

What you've encountered is something called a multiple-valued function. They're an important part of complex analysis. You're correct in your assumptions that a complex number can be represented by multiple "rotations" of the exponential representation.

The solution to your problem is that when you rotate through the full argument range and come back around, you're on a different level of the Riemann Surface. The points at which you cross onto the next level are known as branch cuts or branch points. They're a very important part of complex analysis and they very much govern how you do certain things, like contour integrals.

The solution is when you are doing something like what you're doing, you generally work on the principle branch, which is generally defined as [imath]\theta \in [0,2\pi)[/imath].

- MartianInvader
**Posts:**808**Joined:**Sat Oct 27, 2007 5:51 pm UTC

### Re: i^i

Raising complex numbers to imaginary powers is typically a multi-valued function. We define x^y as e^{y ln x}. e raised to powers is a genuine function, but ln isn't, exactly. ln(1) could be 0, but it could also be 2 \pi i, 4 \pi i, -2 \pi i, etc. In general, asking for the natural log of a number gives you an infinite set of answers, each offset from the next by 2 \pi i.

We typically talk about choosing a branch of the natural log function - that is, choosing a certain range of imaginary values we want our answer to have. For example, we could ask that all the values fall between -\pi and \pi, and then we'll get a ln function that's not multi-valued. Once we pick a branch of the natural log function, we can then use that branch to make well-defined exponentiation - x^y is e^{y ln x} for that particular branch of ln.

Edit: Gorcee beat me to talking about branches, but oh well.

We typically talk about choosing a branch of the natural log function - that is, choosing a certain range of imaginary values we want our answer to have. For example, we could ask that all the values fall between -\pi and \pi, and then we'll get a ln function that's not multi-valued. Once we pick a branch of the natural log function, we can then use that branch to make well-defined exponentiation - x^y is e^{y ln x} for that particular branch of ln.

Edit: Gorcee beat me to talking about branches, but oh well.

Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

### Re: i^i

As an aside, this is one of the main reasons that I think it is better to write exp(x), rather than eVicious Chicken wrote:Take the beloved identity [imath]e^{2\pi i} = 1[/imath]. We can use it to write

[math]e = e^{2\pi i}\cdot e = e^{2\pi i + 1}[/math]and then reapply this to itself to get

[math]e^{2\pi i} = (e^{2\pi i + 1})^{2\pi i} = e^{-4\pi ^{2}}e^{2\pi i}[/math]

^{x}. The latter notation makes it look like exp is a special case of x

^{y}when in fact it is better not to think of it that way, at least in the context of complex analysis.

Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

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