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2 Answers

Euler's Formula:

e^(ix) = cos(x) + i*sin(x)

If x is Pi/2 then:

e^( i * Pi / 2 ) = cos( Pi/2 ) + i * sin ( Pi/2 )

e^( i * Pi / 2 ) = 0 + i

e^( i * Pi / 2 ) = i

raise both sides to the power i like this:

( e^( i * Pi / 2 ) ^ i = i ^ i

simplify

e^( i * i * Pi / 2 ) = i ^ i

e^( -1 * Pi / 2 ) = i ^ i

e^( -Pi/2 ) = about 0.20788, so you could say i ^ i = about 0.20788, but that's not all

the reason Pi/2 worked is because cos(Pi/2) = 0 and sin(Pi/2) = 1

we can go around a full circle and see that cos(5Pi/2) = 0 and sin(5Pi/2) = 1

9Pi/2 would also work, 13Pi/2 would work, and so on

the form ends up being (4n+1)Pi/2, where n is any integer (works for negative and positive integers)

Answer:  e^( -(4n+1)Pi/2 ) where n is any integer

Extra reading:  http://www.math.hmc.edu/funfacts/ffiles/20013.3.shtml

by Top Rated User (103k points)

i=cos(π/2)+isin(π/2)=e^(iπ/2).

i^i=(e^(iπ/2))^i=e^(-π/2)=0.20788 approx.

This is the most basic solution, because 1=sin(π/2)=sin(2π+π/2)=sin(5π/2)=... leading to more non-complex solutions. (e^(-5π/2)=0.000388 approx).

by Top Rated User (825k points)

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