Question #615a2

1 Answer
Oct 7, 2016

#i^i = e^(-pi/2)#

Explanation:

Using Euler's formula , #e^(itheta) = cos(theta)+isin(theta)# we have

#i = 0 + i*1 = cos(pi/2)+isin(pi/2) = e^(ipi/2)#

Substituting that in for the base in in our problem, and applying the property #(x^a)^b = x^(ab)#, we get

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

Thus, we find the interesting result that #i^i# is a real number.