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?
$\endgroup$ 14 Answers
$\begingroup$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 $x$.
$\endgroup$ 1 $\begingroup$The property :
$$\sqrt{x} = x^{1/2}$$
only applies for $x\in \mathbb R$.
$\endgroup$ 5 $\begingroup$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 $\pm1$.
$\endgroup$ $\begingroup$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.
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