The meaning of i^i

Imaginary numbers, like most things at the peripherals of our grasp of reality, interest me, since they appear to have their own rules at times. There was a video blog some time ago that tried to reckon with \large i^i and considered what it equalled.

One calculation I observed came from De Moivre’s identity, \large e^{i\theta} = \cos\theta + i \sin\theta. If you let \theta = \frac{\pi}{2}, you get e^{i\frac{\pi}{2}} = 0 + i = i.

Now raising both sides of the equation e^{i\frac{\pi}{2}} = i to the power i, you obtain the result: e^{i\cdot i\frac{\pi}{2}} = i^i, and, since i\cdot i = i^2 = -1, you get: e^{\frac{-\pi}{2}} = i^i.

This is a notable equation because the left hand side are all real numbers, and the right hand side are all complex. The result of the right-hand side is about 0.20788.

But did this crack that mystery? Well, if you consider the origin of the derivation was from De Moivre’s identity, then it would just compel one to think that not only is \frac{\pi}{2} a solution, but all angles co-terminal to it. i^i therefore has infinitely many answers if we are to rely on De Moivre’s Theorem.

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I am Paul King, a math and science teacher. I help maintain the MIT FAQ Archive along with Nick Bolach. I am also the maintainer of the FAQ for

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