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:



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


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


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


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


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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