What is the square root of $i^4$?

Question

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$$\sqrt{(i^2)^2} = \left|i^2\right| = \left|-1\right| = 1$$

Answer 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.

Answer 2

I think it should be $-1$

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

Answer 3

There is no “the” square root of a complex number, because there is no square root function definable as continuous on all of $\mathbb C$. The complex number $i^4=1$ has two square roots, and in the complex domain there’s (only a) little to help us choose one of them over the other. If you’re only in the real domain $\mathbb R$, then a nonnegative number has a preferred square root, the positive one; but once you jump up to the complexes, the story gets much more confusing.

Answer 4

$i^4=1$, and $1$ has two square-roots, $1$ and $-1$. However when you say the square-root, I cleverly deduce that you mean the principal square-root. By the usual conventions that is $1$.