I want to know (i)^i?

Favorite Answer

Use Euler's theorem:

e^(ix) = (e^x)^i

By Euler's Theorem, e^(ipi/2+2pi*k)=i

So i^i = e^(-pi/2)*e^(2ipi*k) = e^(-pi/2)...Show more

In polar form:

i = cos(π/2) + i sin(π/2) ==> i = e^(iπ/2).

Applying exponential rules:

i^i
=> [e^(iπ/2)]^i
= e^(iπ/2 * i)
= e^(i * i * π/2)
= e^(i^2 * π/2)
= e^(-π/2).

I hope this helps!...Show more

i^i = e^(-pi/2)...Show more