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Prove that i^i is real.

Here we are examining the imaginary number ‘i’, defined as the square root (sqrt) of ‘-1’.

We begin by using Euler’s identity:

e^(i*π)+1=0

e^(i*π)=-1

Since sqrt(x) is the same as x^(1/2):

(e^(i*π))^(1/2)=sqrt(-1)

Using (a^b)^c=a^(b*c) and the definition of i=sqrt(-1):

e^(i*π/2)=i

Then to achieve i^i as specified in the question:

(e^(i*π/2))^i=i^i

Using (a^b)^c=a^(b*c) again:

e^(-π/2)=i^i

Since the LHS has no imaginary part, the RHS is also real. We have proven that i^i is real and equal to e^(-π/2).

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