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

• 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 9 '13 at 8:12
• @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 Jun 9 '13 at 8:34

The unwinding number is defined by $$\ln(e^z) = z + 2 \pi i \mathcal{K}(z).$$ Obviously, $\mathcal{K}(z) \in \mathbb{Z}$.
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 $0$.
• 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 7:34