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For real numbers, $\sqrt{a\cdot b} = \sqrt{a}\cdot\sqrt{b}$ only if at least one of $a$ or $b$ is greater than $0$.

What does the corresponding rule look like for complex numbers? I understand that complex numbers aren't really comparable to each other (or $0$).
For $a = 2+3i$ and $b=3+2i$, $\sqrt{a\cdot b} = \sqrt{a}\cdot\sqrt{b}$ is true.
However, it clearly doesn't work for $a = -2-3i$ and $b=-3-2i$

Angelo
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patentfox
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    For non-negative real numbers $x$, $\sqrt x$ usually denotes the non-negative square root, but such square roots aren’t uniquely defined for complex numbers – J. W. Tanner Apr 23 '21 at 22:39
  • @J.W.Tanner Does that mean a statement like $\sqrt{-4} = 2i$ incorrect/incomplete? $\sqrt{-4}$ could have been $-2i$ as well, as there isn't a well defined way to compare $2i$ or $-2i$ to 0. – patentfox Apr 23 '21 at 22:46
  • I agree with that – J. W. Tanner Apr 23 '21 at 23:06
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    Yeah. $i = \sqrt{-1}$ and $\sqrt{-4} = 2i$ is really one of those irritatingly incorrect statements. $\sqrt{}$ is not really a very usefull or meaningful symbols. ... But this rule is always true. If $\alpha^2 = a$ and $\beta^2 = b$ then $(\alpha\beta)^2 = ab$. Could be the general rule. Or $\sqrt{a}\sqrt{b} = \pm \sqrt{a}\sqrt b$ if we have a way of designating what we mean by "the" square root. – fleablood Apr 24 '21 at 01:19
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    When discussing complex numbers, some authors use the $\sqrt{\quad}$ symbol to denote the principal square root. This usage is fine (and handy) if the convention is made explicit. (Then it must be noted that if the bases aren't real positive numbers, the usual exponential laws may not hold.)$\quad$Going off on a tangent: when $n\in\mathbb Z$ and $w,z\in\mathbb C\setminus{0},$ it is true that $(wz)^n = w^n z^n.$ – ryang Apr 24 '21 at 06:01

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