Complex Exponentiation is Complex

Let's say we want to evaluate

$\frac{ (-1) ^ {-i} } { i ^ i } = ( a \sqrt{a} ) ^ \pi.$

What are wrong with the following methods?

3rd Method:

Raise both sides to the power of 4

$\large \frac { ((-1)^4)^{-i}}{(i^4)^i} = a^{6\pi}$, then $\frac {1^{-i}}{1^i} = 1 = a^{6\pi}$

With that, we have $a = 1$

4th method:

Raise both sides to the power of 3

$\large \frac { ((-1)^3)^{-i}}{(i^3)^{i}} = \frac {((-1)^{-1})^{i}} {(-i)^i } = \frac {(-1)^i}{(-i)^i} = \left ( \frac 1i \right )^i = (-i)^i = a^{9\pi /2}$

$(-i)^{-i} = a^{-9\pi /2}$ or $e^{-\pi /2} = a^{-9\pi /2}$, then $a = e^{1/9}$.

5th Method:

Raise both sides to the power of 5

$\large \frac { ((-1)^5)^{-i}}{(i^5)^i} = a^{15\pi /2}$

$\large \frac {(-1)^{-i}}{i^i} = \left ( \frac {-1}{i} \right )^i = i^i = e^{-\pi /2} = a^{15\pi /2}$

$a = e^{-1/15}$.

6th Method:

Take square root to both sides of the equation, with numerator of fraction becomes $(-1)^{-i/2} = ( \sqrt{-1} )^{-i} = i^{-i}$

$\large \frac {i^{-i} }{i^{i/2}} = a^{3\pi /4}$

$i^{-3\pi /2} = a^{3\pi /4}$

$i^{-1} = a^{1/2}$

$-i = a^{1/2}$

$a = (-i)^2 = -1$.

7th Method:

Knowing that $i^2 = -1$, we could have $(-1)^{-i} = i^{-i/2}$

$\large \frac { i^{-i/2} }{ i^i} = i^{-3i/2} = a^{3\pi /2}$

$i^{-i} = a^\pi, i^i = a^{-\pi}, e^{-\pi /2} = a^{-\pi}$

$a = e^2$.

8th Method:

Knowing that $i^2 = -1$, we could have $i^i = (-1)^{i/2}$

$\large \frac {(-1)^{-i} }{(-1)^{i/2}} = (-1)^{-3i/2} = a^{3\pi /2}$

$(-1)^{i} = a^{\pi}$

$(e^{i \pi} )^{i} = a^\pi$

$a = e^{i^2} = e^{-1}$.

9th Method:

Like the 8th Method

$(-1)^{i} = a^{\pi}$

$i^{i/2} = a^{\pi}$

$i^i = a^{2 \pi}$

$e^{-\pi /2} = a^{2 \pi}$

$a = e^{-1/4}$.

Original problem. See the first 2 methods there.