Csdrhrt

lets look at the below derivation,

$(x)^frac{a}{b} = (x^a)^frac{1}{b}$

squaring both sides,
$(x^frac{a}{b})^2 = ((x^a)^frac{1}{b})^2$
$(x^frac{a}{b})^2 - ((x^a)^frac{1}{b})^2 = 0$
$(x^frac{a}{b} - (x^a)^frac{1}{b})(x^frac{a}{b} + (x^a)^frac{1}{b}) = 0$

hence we got two solutions of $(x)^frac{a}{b}$ as ,

$(x)^frac{a}{b} = (x^a)^frac{1}{b}$ and $(x)^frac{a}{b} = -((x^a)^frac{1}{b})$

if in the equation $(x)^frac{a}{b} = (x^a)^frac{1}{b}$ instead of squaring we powered by 4 both sides, then the 4 solutions of $(x)^frac{a}{b}$ are,

$(x)^frac{a}{b} = (x^a)^frac{1}{b}$ , $ -((x^a)^frac{1}{b})$ , $iota(x^a)^frac{1}{b}$ and $-iota(x^a)^frac{1}{b}$

and if in the equation $(x)^frac{a}{b} = (x^a)^frac{1}{b}$ instead of powering by 4 we powered by 8 both sides, then the 8 solutions of $(x)^frac{a}{b}$ are,

$(x)^frac{a}{b} = (x^a)^frac{1}{b}$ ,$-((x^a)^frac{1}{b})$ , $iota(x^a)^frac{1}{b}$ , $-iota(x^a)^frac{1}{b}$ , $sqrtiota(x^a)^frac{1}{b}$ , $-sqrtiota(x^a)^frac{1}{b}$ , $sqrt-iota(x^a)^frac{1}{b}$ and $-sqrt-iota(x^a)^frac{1}{b}$

as $iota = sqrt(-1) = (-1)^frac{1}{2} = (-1)^frac{1*2}{2*2} = (-1)^frac{2}{4} $

now if we take, $x=-1 , a=2 , b=4$ ,

and solving for $(-1)^frac{2}{4}$ using the above 8 derived solutions,

$(-1)^frac{2}{4} = ((-1)^2)^frac{1}{4} = 1^frac{1}{4} = 1$
$(-1)^frac{2}{4} = -((-1)^2)^frac{1}{4} = -(1)^frac{1}{4} = -1$

and on keep placing in the remainng 6 solutions we further get some weired answers.

but $sqrt(-1)$ is actually equal to $iota$ then how it can be equal to $pm1$