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Author Topic: Debunking \(\sqrt{ab} \neq \sqrt{a}\sqrt{b}\)  (Read 1944 times)  Share 

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RuiAce

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Debunking \(\sqrt{ab} \neq \sqrt{a}\sqrt{b}\)
« on: December 20, 2018, 08:37:30 pm »
+10
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.)
« Last Edit: December 20, 2018, 10:57:00 pm by RuiAce »

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Re: Debunking \(\sqrt{ab} \neq \sqrt{a}\sqrt{b}\)
« Reply #1 on: December 20, 2018, 08:50:18 pm »
0
I just opened this thread accidentally and my self esteem plummeted.
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Re: Debunking \(\sqrt{ab} \neq \sqrt{a}\sqrt{b}\)
« Reply #2 on: December 20, 2018, 09:17:29 pm »
+2
I just opened this thread accidentally and my self esteem plummeted.

I'm not the most mathsy person here so hopefully someone else can help you more, but I promise this isn't as far from your understanding as you might think and here's my somewhat-simplistic attempt to explain:

If you want to describe a line pointing to a point on the cartesian plane, you could say what co-ordinates it points to (x,y). You could also describe the magnitude of the line ( |line| ) and the angle it makes with the positive direction of the x-axis.   With what Rui is describing, the horizontal axis of the graph is the real component  Re(z) and the vertical axis is the imaginary component Im(z). Think of the argument as being the angle between z and the positive directional of that horizontal axis. z = |z| (cos(arg(z)) + isin(z)) is just another way of describing the point z rather than using (Re(z),Im(z)) co-ordinates.


i is defined by i^2 = -1  (i would be straight up on the graph we're using  i = 1(cos(pi/2) + isin(pi/2)    or you could represent it as (1,0) ).

With this knowledge in mind, try working through Rui's work again

Hopefully this helps, but don't worry if it doesn't :)

lzxnl

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Re: Debunking \(\sqrt{ab} \neq \sqrt{a}\sqrt{b}\)
« Reply #3 on: December 20, 2018, 09:55:48 pm »
+6
I'll summarise one of RuiAce's arguments here.

Let us, for the time being, accept that a complex number can be represented as a vector with a direction and a length. The direction is specified as an angle measured anticlockwise from the positive x axis (the argument). Then, complex multiplication ends up doing two things: you multiply the lengths of the two vectors, and add the two angles. Thus, squaring a complex number squares the length and doubles the angle. How should we define the square root of a complex number then? Sounds obvious; square root the length (which is a non-negative real so nothing fancy here) and halve the angle, right? This is RuiAce's principal square root.

By convention, the angle made by a complex number with the x axis (argument) takes values from -pi to pi; if outside this interval, add or subtract multiples of 2pi until you're inside this interval. So, our square root can only have angles from -pi/2 to pi/2. Uh oh. We can't output every complex number with a square root. Can't output a number with angle 2pi/3, for instance. If you tried the number with angle 4pi/3, well firstly you'd have to put this angle into the region -pi to pi to give you -2pi/3, and then halving this angle gets you -pi/3, not 2pi/3.

To make this even worse, we CAN output every complex number as a product of two square roots. Each square root has an angle from -pi/2 to pi/2, so with two such square roots, we can add the angles to get an angle from -pi to pi, which is the whole complex plane. So, sqrt(a) sqrt(b) can give you all the angles in the entire complex plane, but sqrt(ab) cannot. This is why they can't be the same.
« Last Edit: December 21, 2018, 12:11:35 am by lzxnl »
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Re: Debunking \(\sqrt{ab} \neq \sqrt{a}\sqrt{b}\)
« Reply #4 on: December 21, 2018, 02:52:47 am »
+2
I really love this thread. This discussion is amazing! :D

Just want to add some of my own thoughts. I didn't do HSC, so I might be wandering too far out of the course, but I think some university level discussion doesn't hurt ;)

First, I want to address this:
...
If \(\arg z\) denotes the principal argument of \(z\), it's not necessarily true that \(\boxed{\arg z + \arg w = \arg(zw)}\).
...
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.

Despite apparently not being in the course, I think it needs a little bit more discussion because it seems to ironically be causing some confusion.

The complex argument, \(\arg(.)\), in fact, is a multi-valued function. There are infinitely many arguments for all non-zero complex numbers due to the periodicity of sine and cosine. And so, to form a single-valued function, we define the principal argument, \(\text{Arg}(.)\), such that it lies in the interval \((-\pi,\ \pi]\).

Note: I'm going to use uppercase A in principal argument despite knowing Rui said it's not in the HSC course simply because I can't break a habit.

I think it needs to be made clear that the identity here needs to be taken with the understanding that we're doing arithmetic over \(\text{mod }(-\pi,\ \pi]\), and so in full should be written as: \(\boxed{\text{Arg}(z)+\text{Arg}(w)\equiv\text{Arg}(zw)\mod(-\pi,\ \pi],\quad z,w\in\mathbb{C}\setminus \{0\}}\).

Essentially, I just don't think it's 'right' to say  that the equation isn't necessarily true when in reality, it's given with a particular inflection.

And like Rui said, we can avoid issues by using the \(\arg(.)\) function. Essentially, its identity holds regardless since \(\arg(z)\) denotes an infinite set of numbers, mainly: \(\arg(z)=\left\{\text{Arg}(z)+2\pi k\mid k\in\mathbb{Z}\right\}\).

Anyway, moving on! Note: the following would be outside the HSC course, but nevertheless, I think it's interesting.

Let's for a minute look at the real numbers. When taking square roots over \(\mathbb{R}\), it should be noted that it's possible to choose a "standard square root" (or "principal root") so that its function is continuous. (We decided on the positive numbers, so, \(\sqrt{4}=2\), not \(-2\)). Continuity actually allows for the property \(\sqrt{ab}=\sqrt{a}\sqrt{b}\)!!

So, all that remains is to prove that it is impossible to pick a continuous "standard square root" function over \(\mathbb{C}\). Obviously this is well beyond highschool knowledge, but for those interested: you could proceed by assuming, to the contrary, that there exists a function \(f\), continuous on \(\mathbb{C}\) such that \((f(z))^2=z\). To start, let \(z=re^{i\theta}\in\mathbb{C}\setminus\mathbb{R}\) with \(\theta\in (-\pi,\ \pi)\setminus\{0\}\), let \(f_0(z)=\sqrt{r}e^{i\frac{\theta}{2}}\), and consider the function \(f(z)/f_0(z)\).

So, we clearly run into some issues if it's not possible!

Let \(z=r_1e^{i\theta_1}\) and \(w=r_2e^{i\theta_2}\), where \(-\pi<\theta_1,\theta_2\leq \pi\).

So, thanks to the discontinuous nature of the square root function, in general, \(\sqrt{zw}\neq\sqrt{z}\sqrt{w}\), with equality holding only when \(-\pi<\theta_1+\theta_2\leq\pi\) (which Rui discussed indirectly). Woo!

Should note: although I took a more technical approach, I actually more enjoy Rui's intuitive approach to this idea about accessing points in the complex plane :)

...
Complex analysis leads to more powerful results, but getting there is much more of a trek.
...
I couldn't agree more. I think complex analysis is so interesting. I can't wait to formally take it at uni!!
« Last Edit: December 21, 2018, 03:03:54 am by dantraicos »
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lzxnl

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Re: Debunking \(\sqrt{ab} \neq \sqrt{a}\sqrt{b}\)
« Reply #5 on: December 22, 2018, 12:49:04 am »
+2
Should note: although I took a more technical approach, I actually more enjoy Rui's intuitive approach to this idea about accessing points in the complex plane :)
I couldn't agree more. I think complex analysis is so interesting. I can't wait to formally take it at uni!!
Hahahaha complex analysis is great. At the cost of having more requirements for differentiability, you get so much more from it. No more infinitely differentiable functions that are not equal to their Taylor series. Any function differentiable once in a neighbourhood around any point is differentiable infinitely many times about that point and admits a Taylor series. Isn't that beautiful?
Bounded non-constant functions can't be differentiable everywhere. Every differentiable everywhere function (that's not constant) blows up in some direction.

And even better, you can now solve sin(x) = 2.
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