i^i

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November 1st 2005, 12:14 AM
hoeltgman

i^i

I read somewhere that:
$i^i = e^{-\frac{\pi}{2}}$
Where i is the complex unit number
There wasn't any further explanation on this. Only a short developpment.
I don't have any problem with the developpment. But what puzzles me is the fact the the complex power of a complex number is in fact a real number.
Does anybody have a logical explanation for this. I think it's quite a unique phenomenon.
If you take a number out of R and take its power, it will be a number in R again and not Q.
$\pi^\pi \text{\; is in R for example}$
November 2nd 2005, 09:19 AM
Jameson

Check out this mathworld site: http://mathworld.wolfram.com/i.html

Basically you start out with the Euler formula $e^{ix}=\cos(x)+i\sin(x)$ . When you test the value $\frac{\pi}{2}$ , you simply get i, so now we can say that $i=e^{i\frac{\pi}{2}}$ . So, $i^i=(e^{i\frac{\pi}{2}})^i$ . And simple power rules show from here that it reduces to $e^{-\frac{\pi}{2}}$

It makes sense after I go through all of the steps, but I agree I don't see a logical explanation of why this works the way it does. The mysteries of math.

Jameson
November 3rd 2005, 01:50 PM
CaptainBlack

Quote:

Does anybody have a logical explanation for this. I think it's quite a unique phenomenon.
If you take a number out of R and take its power, it will be a number in R again and not Q.

How about (-1/2)^(-1/2)?
November 3rd 2005, 03:15 PM
Jameson

Quote:

Originally Posted by CaptainBlack

How about (-1/2)^(-1/2)?

$\left(-\frac{1}{2}\right)^{-\frac{1}{2}}$ ... touché!

Jameson
November 7th 2005, 08:35 AM
Rebesques

Quote:

If you take a number out of R and take its power, it will be a number in R again and not Q.

You mean C, and you are altogether right.

Consider the function f(z)=z^i. Its graph lies in C ^2. And, it happens to cross the real axis for z=i. Nothing extraordinary here, just a complex stranger, coming from C to the house of the reals for tea. and returning to his imaginary field. :(
November 16th 2005, 07:21 PM
ThePerfectHacker

i^i=e(-pi/2) derived from Euler's Formula:
e^(ni)=cos(n)+isin(n).
November 28th 2005, 09:07 PM
lewisje

Here's another way to think of this:

For all complex a and b, a^b:=e^(b*Ln(a)), where ln(a)=ln|a|+i*Arg(a)+2n*i, n is an integer, and the principal value, Ln(a), is the value for which n=0.

Anyway, Ln(i)=ln(1)+i*Pi/2=i*Pi/2, so i^i=e^(-Pi/2)

As for an example of irrational a and b such that a^b is rational, just consider (2^Sqrt(2))^Sqrt(2)=4.