Does $\sqrt{i^4} = i^2$?

Question

I'm assuming it doesn't, because if it did, then $1 = \sqrt{1} = \sqrt{i^4} = i^2 = -1$. In general, does $\sqrt{x^4} = x^2$?

Answer 1

If $a\in[0,+\infty)$, then $a$ has one and only one real non-negative square root, which we usually denote by $\sqrt a$. Of course, $-\sqrt a$ is also a square root of $a$.

In general, $z\in\mathbb{C}\setminus\{0\}$ then $z$ has two square roots, and if you write $\sqrt z$ it is, in general, not clear which one you have in mind. So, it is perhaps better if you would not use that expression.

Having said this, yes, $i^2(=-1)$ is a square root of $i^4(=1)$.

Answer 2

In general, does $\sqrt{x^4} = x^2$

In $\mathbb R$ yes, because $x^2\ge0$ and $(x^2)^2 = x^4$. So $x^2$ is the positive square root of $x^4$.

On the other hand, in $\mathbb C$, the $x\to\sqrt{x}$ operator is not well define (as explained in other answers), so I'll avoid using it

Answer 3

In general,

$$\sqrt{x^2} = \pm x$$

so technically,

$$\sqrt{i^4} = \pm i^2$$

so what you gave is one of the possible solutions.