SOLUTION: FIND THE VALUES OF i^-i

Question 932083: FIND THE VALUES OF i^-i

Found 2 solutions by MathLover1, rothauserc:
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!
The imaginary number is defined solely by the property that its square is :

With defined this way, it follows directly from algebra that and are both square roots of .
Making use of Euler's formula, is
where is element , the set of integers.
The principal value (for ) is or approximately .
in your case, we have:

The principal value (for ) is or approximately .
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
i^-i = 1 / i^i
we start with Euler's formula
e^(i*pi) + 1 = 0 which implies
e^(i*t) = cos(t) + (i*sin(t))
now we can write
e^(i*(Pi/2)) = cos(Pi/2) + i*sin(Pi/2) = i
now raise both sides of = to the ith power
e^(i*i*Pi/2) = i^i
e^(-Pi/2) = .20788
therefore
(1/(i^i)) = (1 / .20788) = 4.810467577
now note that e^(5i*Pi/2)=i
i^i has many possible values and i^i is a multi-valued function

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