How can i^i be a real number?

Question

I read in a book that if i represents the imaginary number, square root of -1, then i^i is a real number. Why is this?

Answer 1

The "imaginary" (a most misleading term!) i can be represented in the complex plane by
e^((4n+1)*i*π/2) where n is any integer.

[ For simplicity, we might define e^ix as
cos x + i*sin x, and it follows the normal rules for operating with exponents]

Then i^i
= [e^((4n+1)*i*π/2)]^i
and since we multiply the exponents when raising a power to a power, and i*i = -1, this becomes
e^(-(4n+1)π/2).

In the simplest case, n = 0, we get the value
e^(-π/2), and that's the only result I've ever seen anyone else obtain, but as you can see there is an infinity of other possibilities, such as
e^(-5π/2), e^(-9π/2), e^(3π/2), e^(7π/2), e^(-21π/2), e^(15π/2)

All of these values are different from each other, so I question whether the expression i^i has any validity at all since there's an infinite number of possible values for it.

Answer 2

Complex numbers are often best handled as complex exponentials of form re^(iθ), where the correspondence is given by
a + ib = re^(iθ)
<=> a = r cos θ, b = r sin θ
=> r = √(a^2 + b^2), tan θ = b/a
(Similar to polar coordinates; complex numbers can be viewed as the (x, y) plane where y-values are the imaginary part.)
This has the advantage that multiplication is simple in this form: r1 e^(iθ1) . r2 e^(iθ2) = r1 r2 e^(i(θ1 + θ2)). (see De Moivre's Theorem.)

For the complex number i, a = 0 and b = 1, so we get r = 1 and (sin θ = 1 and cos θ = 0) => θ = π/2.
So i = e^(iπ/2).

Now i^i = (e^(iπ/2)) ^ i
By the rules of powers this is equal to e^((iπ/2)(i))
= e^(i^2 π/2)
= e^(-π/2)
so it is a real number.

As to how an imaginary number to the power of an imaginary number can be a real number, really I think it's not much more strange than a transcendental number to the power of another transcendental number being an integer. But this is a familiar situation: for example, e^(ln 2) = 2. Both situations have a similar explanation: if you look at e^x, as you move x along e^x will hit integers once in a while. Similarly, if z and w are complex numbers, as you vary w you will get a real number for z^w once in a while.

Answer 3

the answer is -4. Having in ordinary words a negative signal outdoors the novel image ability an similar because the range being -a million. the three tells you that you want the cubed root of the radicand. 4^3 = sixty 4 4 * -a million = -4

Answer 4

Try writing i as e^(i * pi / 2). Then take this to the i'th power: e^(i * i * pi / 2) = e^(-pi / 2), which is a real number.

Answer 5

heyyy

i=e^ipiby2;
as e^io=cos0+isino;(o=theeta)
therefore i^i=e^i(piby2)*i;(as i.i=1);
=e^-(piby2)
hence this is a rael number