When does $\sqrt{wz}=\sqrt{w}\sqrt{z}$?

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 $\sqrt{wz}=\sqrt{w}\sqrt{z}$?

Solution 1:

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.

Solution 2:

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.