How do you raise a number to a complex power?

[Splanky]

I’m reading a book on the Riemann Zeta function, and it is just getting into the idea that it deals with rasing numbers to complex powers, i.e. X[sup]a+bi[/sup]. It skips over the concept of how to do that, assuring the reader that it can be done but requires advanced analysis. The book is obviously for the layman, which is ok for me because I am still in high school and I have a limited grasp of advanced math.

So, can anyone explain to me how it can be done? Be gentle with me, please.

[Mbossa]

Well, to start off with, you can use one of the laws of exponents to make things a little easier:

x[sup]a+bi[/sup] = x[sup]a[/sup]x[sup]bi[/sup]

x[sup]a[/sup] can be worked out as usual. Now you just need to work out x[sup]bi[/sup]. To do this, you can use a useful conversion:

a[sup]b[/sup] = e[sup]b ln(a)[/sup]

So x[sup]bi[/sup] = e[sup]bi ln(x)[/sup]

Now we can use Euler’s identity:

e[sup]ti[/sup] = cos(t) + i sin(t)

Just substitue b ln(x) for t and you’re done.

[ultrafilter]

@Splanky

I’m reading a book on the Riemann Zeta function, and it is just getting into the idea that it deals with rasing numbers to complex powers, i.e. X[sup]a+bi[/sup]. It skips over the concept of how to do that, assuring the reader that it can be done but requires advanced analysis. The book is obviously for the layman, which is ok for me because I am still in high school and I have a limited grasp of advanced math.

So, can anyone explain to me how it can be done? Be gentle with me, please.

It can’t be done gently. You know how there are two square roots of any number, right? Well, in real analysis, you can just gloss over the second root by making a function that pairs up a number with its positive square root.

In complex analysis, you can’t ignore that sort of stuff. It turns out that x[sup]y[/sup] may have an infinite number of values when x and y are complex. From here, things get interesting–and then they get ugly.

But if you’re not interested in doing advanced calculus in C, you can do pretty much what Mbossa said. It works out that the (primary) value of ln(-1) is [symbol]pi[/symbol], so ln([symbol]i[/symbol]) = [symbol]pi[/symbol]/2. That should get you as far as a layman’s understanding.

[j_sum1]

[hijack]
What book ya reading [n]Splanky**? I recently finished Prime Obsession by John Derbyshire. Great intro to Zeta for an amateur like me.
[/hijack]

[Mathochist]

@Splanky

I’m reading a book on the Riemann Zeta function, and it is just getting into the idea that it deals with rasing numbers to complex powers, i.e. X[sup]a+bi[/sup]. It skips over the concept of how to do that, assuring the reader that it can be done but requires advanced analysis. The book is obviously for the layman, which is ok for me because I am still in high school and I have a limited grasp of advanced math.

So, can anyone explain to me how it can be done? Be gentle with me, please.

Well, once you notice that e[sup]x[/sup] obeys the “exponential” rule (e[sup]x[/sup]e[sup]y[/sup] = e[sup]x+y[/sup], (e[sup]x[/sup])[sup]y[/sup] = e[sup]xy[/sup]), and you have an inverse (a = e[sup]ln(a)[/sup]), you can define a[sup]x[/sup] = e[sup]x*ln(a)[/sup]. You define the exponential function as a power series and the natural logarithm the same way, so they’re defined for complex numbers. That’s all there really is to it.

[BobLibDem]

The math explanations have already been made. I would just like to comment that it is so refreshing to see a young person doing something constructive with his mind and not wasting it watching MTV. Congratulations, Splanky.

[Splanky]

Thanks, BobLibDem ! I do love reading, especially non-fiction. Unfortunately, next year’s AP English course is making me read things I ordinarily wouldn’t- Beowulf , Great Expectations , and other British classics. It’s pretty enjoyable, though.

For the record, the book (which I finished a few days ago) was The Riemann Hypothesis- The Greatest Unsolved Problem in Mathematics. A pretty recent book, for such an old problem (2002, I think).