taylor series for i^i?

[Khan]

I'm having some problems expanding i^i, could anyone help? I know it becomes a real number somehow, and I'm familiar with the e^(i * pi) expansion, but is the i^i done in the same way?

[Njorl]

There is a well known expansion for a^x:

a^x=SUM[((alnx)^n)/(n!)]

Just replace a and x with i.

At first glance, it doesn't look real to me, but maybe the sum telescopes.

Njorl

[Soroban]

Hello, Khan!

I'm not sure what you mean by expanding ii,
since it is already a constant.

Using DeMoivre's Theorem (Euler's?): eix = cos x + i sin x,
when x = pi/2, we have: ei*pi/2 = cos(pi/2) + i sin(pi/2) = i

Raise both sides to the power i: ii = (ei*pi/2)i= e-pi/2 = 0.207879576...