What is i^i?

Question

So, you can raise i to any exponent, and you can raise a number to an imaginary power. But, can you raise an imaginary number to an imaginary power? For example, what is i^i?

You're right Puggy... I just went digging in the answers and found it. (I didn't know how to derive this one, honest.)

Well, I see two posters here have the correct answer. Scythian, I just tossed a coin. If you call it correctly, you get the 10 points. :)

Answer 1

This, too, is an old one. We know that, by Euler's formula,

e^(π i) = -1

So, we can do the following steps:

e^((1/2)π i) = i
e^(-(1/2)π) = i^i

The numerical value for e^(-(1/2)π) is 0.2078795...
A more fun variation on this is the expression for π:

π = -√(-1) Log(-1)

Or that since

π i = Log(-1)

it could be said that the circumference of a circle of diameter of imaginary i is Log(-1). Extending this further, we can say that

π i x = Log((-1)^x)

and it turns out that the function Log((-1)^x) is a sawtooth function that ramps from - π i to π i endlessly with a period of 2. A circumference of minus π i? Anybody know what does that mean?

Addendum: If you tossed a quarter, I call heads. Otherwise, I call tails.

Answer 2

This has been asked in the past. The answer is actually a real number. I'm sure soon you'll have an answerer that can elaborate on this.

Answer 3

i^i is -1
for example if you take a square root of a negative number you have to take the negative out before you can take the square root...the symbol you use is i and i = -1 so i^i is -1^-1 which is -1 mulitplyed by -1 which is -1

Answer 4

Taking Euler's formula

e^(ix) = cos(x) + isin(x)

and substituting (pi/2), for x, one arrives at

e^{i(pi/2)} = i

and

ln(i) = i(pi/2)

If both sides of the first equation are raised to the power i, remembering that i^2 = -1, one obtains this identity:

i^i = e^{i ln(i)} = e^{-pi/2} = 0.2078795763...

Answer 5

i^i = 0.207879576

If you start ut of Euler's Formula:

e^(ix) = cos x + i * sin x

Therefore, when you sub in pi/2 for x you have:
e^(i * pi/2) = i

Which can be expressed as ln(i) = i * pi/2

When both sides are raised to the power of i (remember that i^2 = -1) you have:

i^i = e^(i ln(i)) = e^(-pi/2) = ~0.207879576

I hope this helps

Answer 6

i is used to somewhat rationalize square roots of a negative number. i is said to = -1
so i^i would be 1/i

i^i = 1/i

:)

Answer 7

I tried it on my TI-89 calculator and it gave me an error message.

Answer 8

That's an interesting question. I never thought of it.

Answer 9

-1

Since i = sqrt(-1), it follows that i^2 = -1
i^3 = -i
i^4 =1