What is the square root of $i^4$? What is the $\sqrt{i^4}$? 
$i^4$ = $(i^2)^2$ So is $\sqrt{i^4}$ = $\sqrt{(i^2)^2}$ = $i^2$ = $-1$?
Or is $\sqrt{i^4}$ = $\sqrt{1}$ = $1$?
When I plug it into my TI-89 Titanium, I get $1$.
Edit: I now see why $i^4 \neq -1$. $$\sqrt{x^2} = \left|{x}\right| $$ So$$\sqrt{(i^2)^2} = \left|i^2\right| = \left|-1\right| = 1$$

The answer is $1$.  The reason it is not $-1$ is because, to get there, you inherently split up the square root into a product of square roots as
$$\sqrt{(i^2)^2} = \sqrt{i^2} \sqrt{i^2}$$
This is not valid for nonreal numbers.
