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.
I think it should be $-1$

note that $$i=e^{\left(\frac{\pi}2+2k\pi\right)i}$$ $$i^4=e^{4\left(\frac{\pi}2+2k\pi\right)i}=e^{\left(2\pi+8k\pi\right)i}$$
$$\sqrt{i^4}=e^{\frac12\left(2\pi+8k\pi\right)i}=e^{\left(\pi+4k\pi\right)i}=-1$$