3

I understand that when you raise any number $x$ to a power, you multiply $x$ by itself the number of times indicated in the power. However, what happens when $i^i$ is performed? How can a number be multiplied an imaginary amount of times? Wolfram Alpha says that it is equal to $e^{{-\pi}/{2}}$, but how would you arrive at that answer? Any response will be appreciated, thanks!

Micah
  • 36,226
  • 15
  • 79
  • 126
joejacobz
  • 1,161
  • 1
  • 11
  • 28

4 Answers4

5

$$i^i = e^{i\log i} = e^{i(\log |i|+i\arg i)} = e^{i(i\arg i)} = e^{-\arg i} = e^{-\frac{\pi}{2}+2 \pi k} \qquad k \in \mathbb{Z}$$

Argon
  • 24,247
  • 10
  • 91
  • 130
1

Well, in the complex numbers you consider an exponential of a base other than $e$, such as $z^x$, to be: $$z^x := e^{x\log z }$$ So we have: $$i^i = e^{i\log i}$$ But $\log i = i\left(\frac{\pi}{2}+2\pi n\right)$, so we have $$i^i = e^{ii(\frac{\pi}{2}+2\pi n)} = e^{\frac{-\pi}{2} + 2\pi n} ~~~~~~~~~~ n\in \Bbb{Z}$$ Taking $n=0$ gives the value that Wolfram Alpha gave you.

guest196883
  • 5,790
  • 15
  • 42
  • 4
    I think it should be mentioned that $e^{-\frac{\pi}{2}}$ is the *principal value* of the expression, $i^i$ can take infinitely many real values. – L. F. Feb 08 '13 at 02:03
  • Yes, caveat emptor. – ncmathsadist Feb 08 '13 at 02:05
  • One must be careful about complex powers; this is a branch-of-the-log consideration. – ncmathsadist Feb 08 '13 at 02:06
  • Yes, as L.F. said, log i can produce an infinite number of values, and is only a function upon selecting a branch $i(\frac{\pi}{2} + 2\pi n). $ $e^{-\pi/2}$ is the principal value with $n=0$. – guest196883 Feb 08 '13 at 02:08
-1

For example, $$i^i=(\cos(\pi/2)+i\sin(\pi/2))^i=e^{i(i\pi/2)}=e^{-\pi/2}$$

Clayton
  • 23,935
  • 5
  • 49
  • 107
-1

for any complex number $z\in \mathbb C$ it can be written as $z=x+iy$ or in the polar form $z=re^{i\theta}$ where $r=\sqrt{x^2+y^2}$ and $\theta=\arctan(\frac{y}{x})$, so in particular $i=0+i1$ which implies that $r=1$ and $\theta=\frac{\pi}{2}$ Thus, $$i=e^{\frac{\pi i}{2}}$$ which implies that $$i^i=(e^{\frac{\pi i}{2}})^i=e^{\frac{-\pi }{2}}.$$

i.a.m
  • 2,308
  • 11
  • 30
  • @Argon I know that, but it is still true, the argument of $i$ is $\frac{\pi}{2}$ – i.a.m Feb 08 '13 at 02:24
  • @Argon I understand what you mean, but may be I should've just written that $\theta=\frac{\pi}{2}$ without trying to explain it more. – i.a.m Feb 08 '13 at 02:47