What's the square root of i to the power of 4?

Question

This is not a homework question

$\sqrt{i^4} = \sqrt{1} = 1$

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

So what did I do wrong?

Answer 1

The property :

$$\sqrt{x} = x^{1/2}$$

only applies for $x\in \mathbb R$.

Answer 2

Every nonzero complex number has two square roots whose arguments are 180 degrees apart in the complex plane. You have not done anything wrong and have shown that the number $i^{4}=1$ has two square roots namely $\pm1$.

Answer 3

Another fun paradox: $-1 = (-1)^1 = (-1)^{2/2} = ((-1)^2)^{1/2} = 1^{1/2} = 1$.

More seriously, the property $x^{a b} = (x^a)^b$ only applies for positive real $x$.

Answer 4

In your second equation you've used the property $$(x^a)^b = x^{ab}.$$ Generally you should only use this when $x$ is guaranteed to be a positive real number. You can put different conditions on $x$ (e.g. positive), or $a$ and $b$ (e.g. whole numbers) and still have it work out correctly, but you're really asking for trouble.