One of the math teachers asked me for a nice way to explain how to find . One of her students had asked her. What’s funny is that I distinctly remember coming up with this question when I was in high school, and I remembered my solution.

I’ll put it below in case any of you want to try to work it out.

Starting with Euler’s identity:

We note that

So raising both sides of the equation to the 1/2 power yields:

.

Of course this method will then let you find to any power!

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Here’s another cool way using the rotation model (think of multiplying by -1 as a 180 degree rotation on the complex plane; thus a 90 degree rotation is the square root of -1). This lets you build the imaginary number line perpendicular to the real number line. If you extend the metaphor, a 45 degree rotation would be the square root of i. Plot a complex number 1 unit from the origin at a 45 degree angle. Now you have a picture like the unit circle, imposed on the complex plane. Using basic trig, both components of the complex number must be root(2)/2.

OK, but, what is i^i? :-)

@Dan: Totally get it! Thanks.

@Kate: its e^(-pi/2).

Since i=e^(i*pi/2), raise both sides to the ith power. You get i^i=(e^(i*pi/2))^i=e^(-pi/2).

It’s a plain, old, ordinary real number. Sigh.

Both sqrt(2)/2 + i sqrt(2)/2 and -sqrt(2)/2 – i sqrt(2)/2 are square roots of i, aren’t they? Is there a way we decide which one of them is meant by sqrt(i) (just as when we write sqrt(9) we know we mean 3 and not -3), which we could use for the square root of any complex number? Maybe the one whose angle theta is between 0 and 180?

@Matt, I think that it’s a matter of symantics…

Technically, when someone says “the square root of three” there are two answers — and . But that’s just technical.

In everyday speak, when someone says “the square root of three” the mean and when they mean the negative value, they’ll say “negative the square root of three.”

See http://en.wikipedia.org/wiki/Square_root

So I think generally, we can say that . And we designate the negative solution with the negative values…

Does this make any sense?

While the square root of i has only two possible answers, if I’m not mistaken, i^i has infinitely many. For instance, while i = e^{i pi/2}, it also equals e^{9i pi/2} and e^{-7i pi/2} and e^{17i pi/2} and…

Therefore, at the very least, the elements of the set {e^{-(2k+1) pi/2} | k is an integer} are all answers to the question: what is i^i?

Adam is absolutely right. 0.2078…. is just the principal value (k=0 in Adam’s expression).