How do I compute $(-i)^i$?

Question

Could I please get help computing $(-i)^i$? I know the answer is $$e^{\frac{\pi}{2}}e^{-2\pi n}=e^{-2\pi\left(\frac{n-1}{4}\right)}.$$ I can't figure out how to get to this answer.

Answer 1

Since $-i =e^{3\pi i/2+2\pi i n} =e^{-\pi i/2+2\pi i n} $ for any integer $n$ (different $n$s),

$(-i)^i =(e^{-\pi i/2+2\pi i n})^i =e^{i(-\pi i/2+2\pi i n)} =e^{\pi /2-2\pi n} =e^{\pi /2}e^{-2\pi n} =e^{\pi /2}e^{2\pi n} $ with a different n at the end.

Answer 2

Starting with (-i), we can express it as e^(i * (3π/2)) because cos(3π/2) = 0 and sin(3π/2) = -1.

Now, raise this expression to the power of i. Using the properties of exponents, this becomes e^(i * i * (3π/2)).

Since i^2 = -1, we have e^(-3π/2), which is a real number. So, (-i)^i equals a real number, approximately 1.18.