What is i^i?

Question

Please take me through each step, I'm quite good at maths but I don't know that much so explain properly please.
Apparently it's a real number.

Answer 1

I did this just a couple of days ago, and it is indeed a real number. There are a couple of ways of doing this, but this seems to be the neatest.
You start off by saying that there is some complex number z such that e^z=i.
If z = a+ib; then you have (e^a)*(e^ib)=i.
You then use de Moivre's theorem, which states that:
e^i*x = cos(x) + i*sin(x) to get:
(e^a)*(cos(b) + i*sin(b) ) = i.
Since there is no real part on the RHS, there must be no real part on the LHS.
Therefore cos(b)=0. This means that b must equal pi/2 or 3pi/2. However, if it was 3pi/2, then sin(b) would be -1, and then e^a would have to be -1, which is impossible with real a.
This means that b=pi/2 (well actually pi/2 + 2*n*pi; where n is an integer)
So (cos(b) + i*sin(b) ) = i. Therefore e^a=1, so a=0.
This means that z=0 + i*pi/2.
so i=e^z = e^(i*pi/2)
To get i^i, you just raise both sides to the power i:
i^i= (e^(i*pi/2)^i) = e^(i*(i*pi/2)) = e^(-pi/2).
So your answer is
i^i = e^(-pi/2)
or e^(-(pi/2 + 2*n*pi) )

If you aren't familiar with de Moivre's theorem, you can derive it like this.
Let f(x) = cos(x) + i*sin(x). Differentiate it to get:
f'(x) = -sin(x) + i*cos(x) = i*cos(x) + (i^2)*sin(x). Factorize one of the i's:
f'(x) = i*(cos(x) + i*sin(x) ) = i*( f(x) ). So, if you view 'i' as a constant, f'(x) is proportional to f(x), which means f(x) must be in the form:
f(x) = k*e^(i*x). To find the value of k, just substitute in a value for x, say, 0.
f(0) = ke^0 = k; and f(0) = cos(0) + i*sin(0) = 1.
So k=1.
So we now have that:
e^(i*x) = cos(x) + i*sin(x)
Which is de Moivre's theorem.

ps I hope you are familiar with radians and differentiation and all that sort of stuff.

Answer 2

How about

exp(i^2*p/2i) = exp(-pi/2) , which is a real number!

Answer 3

we know e^ix = cos x + i sin x

put cos x = 0 sin x = 1

so x = pi/2

so i = e^ipi/2

raise both to power i

i^i = (e^ipi/2)^i = e^(ipi/2*i) = e^(-pi/2)

Answer 4

i=cos90+isin90 ( by De Moivre's theorem)
=e^(pi*i/2)

i^i=(e^(pi*i/2))^i
=e^(pi/2*i^2)
=e^(-0.5*pi)
=0.207879
which is real

Answer 5

As cos(t)+j*sin(t)=exp(j*t), then j=exp(j*pi/2); j^j=exp(j*j*pi/2)=exp(-pi/2)

Answer 6

i^i = (e^(i*pi/2))^i = e^(i^2*pi/2) = e^(-pi/2) = 0.207879...