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I'm studying a set of lecture notes on complex numbers, which derive the fact that $\sqrt{i} = \pm \frac{1 + i}{\sqrt{2}}$. I'm fine with this result, but they then comment that, knowing this fact, we can deduce that $\sqrt{2i} = \pm (1 + i)$.

I'm very confused by this, because it is my understanding that the usual properties of square roots, such as $\sqrt{ab} = \sqrt{a} \sqrt{b}$ do not necessarily hold for complex numbers. They give an example where attempting to use such properties leads to the absurd conclusion that $i^2 = 1$.

Is there some other way to deduce this fact, or is this one of the rare cases where the square root function works as it does in $\mathbb{R}$?

Cardinality
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    If for this question $\sqrt{\cdots}$ is multivalued, for $a=b=-1$, $\sqrt{ab}$ gives ${-1, 1}$, and $\sqrt a \sqrt b$ also gives ${-1, 1}$. Still the single-valued square-root property may not hold. – peterwhy Jun 20 '23 at 03:55

1 Answers1

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It is true that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ is not necessarily true for complex numbers $a$, $b$. However, it is true that $\sqrt{ab} = \operatorname{sqrt} {(a)} \sqrt{b}$, where $a$ is a nonnegative number and $b$ is a complex number, and $\operatorname{sqrt} {(a)}$ is the unique nonnegative number whose square is $a$. You are invited to verify the claim yourself.

Postscript: $\sqrt{x}$ is multi-valued here.

Chunyang
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