So I was watching the legendary Mr Eddie Woo's videos again and I stumbled on this one. Which baffled me ever since i first saw it in HSC MX2. Eddie Woo presents it in a very nice way to give intuition about why this formula has the
potential to break when you think about it and it's really just a nice application of quadratic applications.
But it only convinces me about the why. Over time, every now and then I kept thinking about
when does the formula break. Because there has to be some reason why it never breaks at all if \(a>0\) and \(b>0\), but just so happens to if we let \(a = b = -1\). So is there any way of capturing when exactly do we get \( \sqrt{ab} = -\sqrt{a}\sqrt{b} \) as opposed to the positive case?
I don't claim that this is the best explanation. It's also pretty advanced at the 4U level - essentially I'm using complex analysis to help describe what's going on. The idea is that if \(\arg z\) denotes the principal argument of \(z\), it's not necessarily true that \( \boxed{\arg z + \arg w = \arg(zw)} \). (To the university students, I'm intentionally avoiding capital-A Arg because this is still in the 4U section, and not an extracurricular post.)
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Again, \(\arg z\) denotes the principal argument of \(z\) here.
We know that for any complex number \(z\) we can rewrite it as
\[ z = |z| \left( \cos (\arg z) + i \sin (\arg z) \right) \]
because this is literally the mod-arg form (AKA polar form). But what do we really mean by \( \sqrt{z} \)? We've established as a convention to let \( \sqrt{1} = +1\) instead of \(-1\), and we've also established as a convention to let \( \sqrt{-1} = +i\) instead of \(-i\). We can also go further and establish that \( \sqrt{i} = \frac{1}{\sqrt2} + \frac{1}{\sqrt2} i \), instead of \( \frac{1}{\sqrt2} - \frac{1}{\sqrt2}i \). But where do these well-established/new conventions come from?
The idea is that not only is there a thing called a principal argument, but there's also a thing called a principal root. The notation \( \sqrt{z} \) is used to denote the
principal square root, and its definition is
\[ \sqrt{z} = \boxed{\operatorname{p.v.} z^{1/2}} = |z|^{1/2} \left( \cos \left( \frac{\arg z}{2} \right) + i \sin \left( \frac{\arg z}{2} \right) \right). \]
The letters "p.v." essentially stand for "principal value".
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Set-up out of the way, we return to our problem. We have a similar definition for \(zw\):
\[ zw = |zw| (\cos (\arg zw) + i \sin (\arg zw)).\]
This now lets us write down a definition for \(\sqrt{zw}\).
\[ \boxed{\sqrt{zw} = |zw|^{1/2} \left( \cos \left( \frac{\arg zw}{2} \right) + i \sin \left( \frac{\arg zw}{2} \right) \right)} \]
and we note that because \( -\pi < \arg zw \leq \pi\), we now have \( \boxed{-\frac\pi2 < \arg zw \leq \frac\pi2} \).
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However we also have similar definition for \(w\):
\[ w = |w| (\cos (\arg w) + i \sin (\arg w)). \]
Consequently
\[\sqrt{w} = |w|^{1/2} \left( \cos \left( \frac{\arg w}{2} \right) + i \sin \left( \frac{\arg w}{2} \right) \right)\]
But now, recalling that \( \operatorname{cis} \theta \operatorname{cis}\phi = \operatorname{cis} (\theta+\phi)\), or alternatively \( e^{i\theta} e^{i\phi} = e^{i(\theta+\phi)}\), we have
\[ \boxed{\sqrt{z}\sqrt{w} = |zw|^{1/2} \left( \cos \left( \frac{\arg z + \arg w}2\right) + i \sin \left( \frac{\arg z + \arg w}2 \right) \right)} \]
Here's the thing. Note that because \( -\pi < \arg z \leq \pi \) and \( -\pi < \arg w \leq \pi\), when adding we actually end up with \( \boxed{-2\pi < \arg z + \arg w \leq 2\pi} \). Consequently, \( \boxed{-\pi < \frac{\arg z+\arg w}{2} \leq \pi} \).
Note therefore that the range of values \( \frac{\arg z + \arg w}{2} \) can take on is not the same as that of \( \frac{\arg zw}{2} \), but rather a larger one!
The working out has shown that \( \arg \left( \sqrt{zw} \right) = \frac{\arg zw}{2} \), so ultimately \(\sqrt{zw}\) can only be a complex number in the first or fourth quadrant, or alternatively a complex number with non-negative real part.
Whereas on the other hand, \( \arg \left(\sqrt{z}\sqrt{w} \right) = \frac{\arg z + \arg w}{2} \), so ultimately \( \sqrt{z}\sqrt{w}\) can be a complex number in any quadrant.
So if \( \sqrt{z}\sqrt{w} \) ends up being in the second quadrant, third quadrant or along the negative real axis, there is no way whatsoever for it to equal to \( \sqrt{zw}\), which is jammed into the other quadrants.
Note that Eddie Woo's video essentially showed that it has to be equal to its negative otherwise, which is why we don't get something berserk like \(\sqrt{zw} = \sqrt{z}\sqrt{w} \times \left(\frac12 + \frac{\sqrt3}2 i \right)\) for example. But there is a bit of intuition behind all of this working out as to why it should be the negative as well. I won't bother discussing that here.
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Note that we can also convince ourselves this is the case by subbing in \(z=-1\) and \(w=-1\). This would give us \( \arg z = \pi \) and \(\arg w = \pi\), so therefore \( \frac{\arg z+\arg w}{2} = \pi\). Consequently, \( \sqrt{z}\sqrt{w} = \cos \pi + i\sin \pi = -1\).
Yet on the other hand, \(zw\) = 1, so \( \arg zw = 0 \) and furthermore \( \frac{\arg zw}{2} = 0\). Consequently, \( \sqrt{zw} = \cos 0 + i \sin 0 \).
The problem of dealing with principal values is perhaps what makes complex analysis much harder than doing calculus over the real numbers. Complex analysis leads to more powerful results, but getting there is much more of a trek. (It also illustrates why I don't like the formula \( \arg z + \arg w = \arg (zw) \) - this can only be used if we're assuming the multi-valued argument, which we just don't do in high school to avoid confusion.)