Question

What is the value of i^i?
Where i=iota, a complex no. (0,1). Please tell!

Answer 1

the answer is \[e^(-pi/2)\]

Answer 2

e raised to power - pi/2

Answer 3

i have just done upto \[e^\ln(i^i) \]

Answer 4

e raised to the power of log of i squared to the base e

Answer 5

now what should i do from here?

Answer 6

\[i = \cos pi/2 + i (\sin pi/2)\]
Therefore
\[\large i = e^{i(pi/2)}\]

Now\[\huge i^i = e^{(i*i(pi/2))} = e ^ {-pi/2}\]