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There exists a unique function $\sqrt{*} : \mathbb{C} \rightarrow \mathbb{C}$ such that for all $r \in [0,\infty)$ and $\theta \in (-\pi,\pi]$ it holds that $$\sqrt{r\exp(i\theta)}=\sqrt{r}\exp(i\theta/2),$$

where $\sqrt{r}$ denotes the usual principal square root of a real number $r$.

Lets take this as our definition of the principal square root of a complex number. Thus $i=\sqrt{-1}.$

Now. We know that, for all positive real $w$ and $z$, it holds that $\sqrt{wz}=\sqrt{w}\sqrt{z}$. We also know that this fails for certain complex $w$ and $z$. Otherwise, we'd be allowed to argue as follows:

$$-1 = i\cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = \sqrt{1}=1$$

My question: for which complex $w$ and $z$ does it hold that $\sqrt{wz}=\sqrt{w}\sqrt{z}$?

Martin Sleziak
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goblin GONE
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    Seems offhand to me (I'm posting from my phone) that it's necessary and sufficient to have all of $-\pi\lt \theta_1, \theta_2, \theta_1+\theta_2\le\pi$. – MJD May 26 '13 at 14:59
  • $i=\sqrt{-1}$ - that is not true. – Bartek Pawlik Jun 09 '13 at 08:12
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    @BartekPawlik, in the context of these particular definitions it is. In particular, since $-1=\exp(i\pi),$ thus $\sqrt{-1}=\exp(i\pi/2)=i$. – goblin GONE Jun 09 '13 at 08:34
  • Going to close a duplicate of this later post because of all the answers there, although the answer here is also extremely interesting and hopefully will be brought to greater attention. – rschwieb Jun 01 '20 at 13:56
  • Does this answer your question? When does $\sqrt{a b} = \sqrt{a} \sqrt{b}$? – Adrian Keister Jun 01 '20 at 17:42
  • @AdrianKeister, I mean, isn't the quality of my question higher than the quality of the linked question? Note that the linked question doesn't define square roots, it just assumes everyone is on the same page, whereas my question carefully ensures that we're all speaking the same language. Also my question as asked first. Additionally, Adayah has written a clear elementary answer that should be useful to a broad audience. – goblin GONE Jun 02 '20 at 10:03
  • @rschwieb, I think this question should be kept and the other one closed. Or maybe they should both be left alone. I think questions are closed too readily. Why not just interlink them? – goblin GONE Jun 02 '20 at 10:04
  • @goblin Both will exist and they are interlinked... so it seems that is in accord with what you just said. I can’t see much of an argument for duping I the opposite direction – rschwieb Jun 02 '20 at 10:44
  • On the quality of the question, I have to say: never have I seen this problem asked so well. Any other time it was "why is this proof of $1 = -1$ wrong" and the asker wasn't even aware that the $\sqrt{z}$ symbol is ambiguous and unclear unless explicitly defined. – Adayah Jun 02 '20 at 17:35
  • @Adayah, thanks! – goblin GONE Jun 04 '20 at 08:15

2 Answers2

6

The solution to this question requires the definition of the unwinding number. Check the paper The unwinding number by Corless and Jeffrey, SIGSAM Bulletin 116, pp. 28-35.

The unwinding number is defined by $$\ln(e^z) = z + 2 \pi i \mathcal{K}(z).$$ Obviously, $\mathcal{K}(z) \in \mathbb{Z}$.

For your question, Theorem 5 is the most relevant one, along with the point 1 in the second list in section 5.2:

  1. $\sqrt{zw}$. By theorem (5c) we would expect this to expand to $$\sqrt{z}\sqrt{w}e^{\pi i \mathcal{K}(\ln z + \ln w)}$$ and this would not simplify further unless the assume system knew that $-\pi < \arg z + \arg w \le \pi$, in which case $\mathcal{K}$ would simplify to $0$.

Read the paper for a deeper insight.

Vedran Šego
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    The punchline is that MJD's comment is right: $\sqrt{zw}=\sqrt{z}\sqrt{w}$ if and only if $-\pi < \arg z + \arg w \le \pi$. Maybe that should be emphasized. – Chris Culter Jul 16 '15 at 07:34
5

I know the question is old, but the accepted answer not only relies on an inactive external resource, but also feels needlessly complicated: neither the complex logarithm, nor the unwinding number have any relevance to the question.

Write $w = r \exp(i\varphi), z = s \exp(i \psi)$ where $\varphi, \psi \in (-\pi, \pi]$. Then the equality holds if and only if either $wz=0$ or $\varphi + \psi \in (-\pi, \pi]$.

Proof: clearly if either $w$ or $z$ equals zero, then both sides of the equality are zero as well, so from now assume $wz \neq 0$. Three cases are possible:

  • $\varphi + \psi \in (-\pi, \pi]$

    Then $wz = rs \exp \big( i(\varphi+\psi) \big)$, so by definition

    $$\sqrt{wz} = \sqrt{rs} \exp \left( i \cdot \frac{\varphi+\psi}{2} \right) = \sqrt{r} \exp \left( i \cdot \frac{\varphi}{2} \right) \cdot \sqrt{s} \exp \left( i \cdot \frac{\psi}{2} \right) = \sqrt{w} \sqrt{z}.$$

  • $\varphi + \psi \in (-2\pi, -\pi]$

    Then $wz = rs \exp \big( i (\varphi+\psi+2\pi) \big)$ and $\varphi+\psi+2\pi \in (-\pi, \pi]$, hence

    $$\sqrt{wz} = \sqrt{rs} \exp \left( i \cdot \frac{\varphi+\psi+2\pi}{2} \right) = \sqrt{r} \exp \left( i \cdot \frac{\varphi}{2} \right) \cdot \sqrt{s} \exp \left( i \cdot \frac{\psi}{2} \right) \cdot \exp(i\pi) = -\sqrt{w} \sqrt{z}.$$

  • $\varphi + \psi \in (\pi, 2\pi]$

    Then $wz = rs \exp \big( i (\varphi+\psi-2\pi) \big)$ and $\varphi+\psi-2\pi \in (-\pi, \pi]$, thus

    $$\sqrt{wz} = \sqrt{rs} \exp \left( i \cdot \frac{\varphi+\psi-2\pi}{2} \right) = \sqrt{r} \exp \left( i \cdot \frac{\varphi}{2} \right) \cdot \sqrt{s} \exp \left( i \cdot \frac{\psi}{2} \right) \cdot \exp(-i\pi) = -\sqrt{w} \sqrt{z}.$$

As $-\sqrt{w} \sqrt{z} \neq \sqrt{w} \sqrt{z}$, the equality does not hold in the last two cases, so the proof is complete.

Adayah
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