what is $\sqrt{(i^4)}$ We know that,

$(a^m)^n =  a^{mn} = (a^{n})^m$

So , 


*

*$\sqrt {i^4} = (i^{1/2})^4 = \left(\pm\frac{1+ i}{\sqrt2} \right)^4 = -1$

*$\sqrt {i^4} = (i^{2}) = -1$

*$\sqrt {i^4} = \sqrt1 = 1$
I think only no $3$ is right. I must have violated some rule in $1$ & $2$. Please let me know.
Maybe $(a^m)^n =  a^{mn} = (a^{n})^m$ is true only for Real Numbers.

$$\sqrt{i^4}=i^2 = -1\tag1$$

Or
$$\sqrt{i^4}= \sqrt{(\sqrt{-1})^4} = (\sqrt{-1})^2 = -1 = i^2 \tag2$$

Or
$$\sqrt{i^4} = \sqrt{i^2\cdot i^2} = \sqrt{i^2}\cdot \sqrt{i^2}= i\cdot i = i^2  = -1\tag3$$
