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I understand that the two square roots of $i$ are

$$\pm {1 \over \sqrt{2}} (1+i)$$

But when asked what is the square root of $i$, is it equally correct to give the answer as

$$\pm {(-1) ^ {1/4}}$$

? And actually, since $\sqrt 2$ is defined to be only the positive root, so square root of $i$ can be just

$${(-1) ^ {1/4}}$$

?

nonopolarity
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2 Answers2

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This is easiest to see in polar form. $i$ is at an angle of $\frac \pi 2$. Its square roots are at angles of $\frac \pi 4$ and $\frac {5\pi}4$. Its fourth roots are at $\frac \pi 8 + k\frac \pi 2$, which do not match at all.

Ross Millikan
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  • sorry, I actually meant $$\pm {(-1) ^ {1/4}}$$ ... I was think of the root of root of $-1$ but wrote it as the root of the root of $i$ – nonopolarity Jan 23 '21 at 16:43
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I'll ignore $0$ throughout. Also, $n$ will be a positive integer.

First, with the more familiar real numbers. When $n$ is even, positive numbers have two distinct $n$th roots and negative numbers have none. When $n$ is odd, all numbers have one $n$th root.

It is quite different for the complex numbers. All non-zero numbers have $n$ distinct $n$th roots.

So, what you say cannot be correct as there will be two square roots but four fourth roots.

A further common problem is talking of the square root, etc. Among the reals, we can choose that the positive square is the principal one and gets to be called the square root. For the complex numbers, we cannot pick a preferred root so neatly and some familiar behaviour, e.g. $\sqrt{x}\times \sqrt{y} = \sqrt{xy}$ might not be true.

badjohn
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