1

As I was going through my dad's college magazine (of 1989), in the mathematics department quiz section, there was a question:

What is $i^i$, where $i=(-1)^{1/2}$, that is, is $i^i$ a real number, an imaginary number, or undefined?

I was suddenly taken aback, I mean how can be $i$ be raised to the power $i$, I thought may be Cartesian representation helps a bit in solving it but it led me nowhere so can anyone help?

Did
  • 279,727
Arnav Das
  • 753
  • What is $i^i$? A nonexistent entity. – Did Sep 06 '15 at 08:21
  • Yeah but the answer says it's a real number !!!!! – Arnav Das Sep 06 '15 at 08:22
  • @Did that's the precise question...... – Arnav Das Sep 06 '15 at 08:23
  • 2
    $i \ne \iota$. ($i$ is i, $\iota$ is iota.) – fkraiem Sep 06 '15 at 08:24
  • 3
    This is one abuse sometimes made, even by mathematicians that should know better, that since $i=e ^{i\pi/2}$, $i^i$ should be $e^{i\cdot i\pi/2}=e^{-\pi/2}$. Only, by the same reasoning, $i=e ^{5i\pi/2}$ hence $i^i$ should also be $e^{i\cdot 5i\pi/2}=e^{-5\pi/2}$... and then chaos ensues. The real trouble is that no continuous function $z\mapsto z^i$ can be defined on the whole complex plane. – Did Sep 06 '15 at 08:25
  • 1
    @ArnavDas Try writing $i=e^{i\pi/2}$ and think about what happens. Also think about the fact that $i=e^{5i\pi/2}$. – Peter Woolfitt Sep 06 '15 at 08:25
  • One can create branches of a multibranch function, and in one of those branches, we get one of those values, @Did. Or we could have the "function" be a set function to begin with, like the arg(Z), which returns all of the arguments possible. – Alan Sep 06 '15 at 09:24
  • 1
    @Alan We all know that, thanks. The problems I have with the multivalued function approach are that (1) I fail to see the advantages, (2) most people using it forget the "multi" part on the road. – Did Sep 06 '15 at 09:32
  • @Did Just look at $z \mapsto i^z$ rather than $z \mapsto z^i$. Then you have entire functions (one for each choice of $\log i$). The trouble goes deeper than that power functions with non-integer exponents aren't entire. – Daniel Fischer Sep 06 '15 at 10:40

0 Answers0