How do I put i^i into a+bi form?

Answer 1

cos(theta)+i sin(theta) = e^(i. theta)

so i = e^(i. pi/2), putting theta=pi/2

i^i = e^(i.i.pi/2) = e^(-pi/2), an astonishingly real number

Answer 2

Just remember that i=exp(i*pi/2) (source: http://mathworld.wolfram.com/i.html), then:

i^i = a + bi
(exp(i*pi/2))^i = a + bi
exp(i^2*pi/2) = a + bi
exp(-pi/2) = a + bi

Then i^i = exp(-pi/2) + 0*i = 0.20788 + 0*i.

I hope this helps.