Quod Erat Demonstrandum

2007/11/24

[AL][PM] Class discussions on complex numbers

Filed under: HKALE,Pure Mathematics — johnmayhk @ 1:21 下午
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We are learning the roots of unity in these F.7 pure math lessons. I randomly gave the following

Solve $x^5 - x^4 + x^3 - x^2 + x - 1 = 0$

Then I wrote

$x^5 - x^4 + x^3 - x^2 + x - 1 = 0$
$1 - x + x^2 - x^3 + x^4 - x^5 = 0$
$\frac{1 - (-x)^6}{1 - (-x)} = 0$ (for $x \neq -1$) – – – (1)

then I jumped to

$x^6 = 1$ – – – (*)

Then I stopped for a while because I had something in my mind:

started with another equation

$x^5 + x^4 + x^3 + x^2 + x + 1 = 0$

we can jump to the same equation $x^6 = 1$, why?

At that moment, Harris claimed that he had a quicker method: “just adding a negative sign".

Well, when solving (*), we have

$x = cis\frac{2k\pi}{6}$ (for $k = 0,1,2,4,5$ < – – – no ‘3’ since $x \neq -1$) is the roots to the original equation.

Then Harris wrote his method (H’s method) on the board as follows.

From (1), we have

$(-x)^6 = 1$
$-x = cis\frac{2k\pi}{6}$
$x = -cis\frac{2k\pi}{6}$ < – – – – this is so-called “adding a negative sign"

Well, it is OK with saying $k = 1,2,3,4,5$ < – – – – no ‘0’ since $x \neq -1$.

Is it a quicker way to obtain solutions, as claimed by Harris? Well, it depends.

But the most important thing is how to explain the solution given by Harris is the same as what I gave.

Of course, we can put different values of $k$ and check them one by one. However, it’s time consuming. We should have a general (or easier) way to show it up.

At that moment, some (or few) students tried to suggest something and think it in details. That is a wonderful moment because interaction, active learning began!

The first thing came to me is trying to see the geometric meaning. But how stupid that I was thinking about vertices of a regular pentagon!! I mistake that there are 5 roots so the roots represent the vertices of a regular pentagon!! No, $cis\frac{2k\pi}{6}$ are representing vertices of a regular hexagon!! A student, Lai, reminded me that there are 6 distinct values generated!!

Then the situation is easy now.

Because the vertices of a regular hexagon is symmetrical about $O$,
that is,
if $cis\theta$ is a root
=> the point representing $cis\theta$ is a vertex P of the regular hexagon
since there is another vertex P’ of the hexagon such that P and P’ are symmetrical about O
and the corresponding complex number represented by P’ is $-cis\theta$
=> $-cis\theta$ is also a root

Don’t understand geometric meaning? Urm, try to use “pure algebra".

For $n$ being a positive even number,

$\cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}$
= $-\cos(\frac{2k\pi}{n} - \pi) - i\sin(\frac{2k\pi}{n} - \pi)$
= $-(\cos\frac{(2k - n)\pi}{n} + i\sin\frac{(2k - n)\pi}{n})$
= $-(\cos\frac{2(k - m)\pi}{n} + i\sin\frac{2(k - m)\pi}{n})$ where $n = 2m$ for some natural numbers $m$.
= $-(\cos\frac{2k'\pi}{n} + i\sin\frac{2k'\pi}{n})$ where $k' = k - m$

Back to our question, $n = 6$, hence $m = 3$; thus, by above,

$cis\frac{2(0)\pi}{6} = -cis\frac{2(0 - 3)\pi}{6} = -cis\frac{2(3)\pi}{6}$
$cis\frac{2(1)\pi}{6} = -cis\frac{2(1 - 3)\pi}{6} = -cis\frac{2(4)\pi}{6}$
$cis\frac{2(2)\pi}{6} = -cis\frac{2(2 - 3)\pi}{6} = -cis\frac{2(5)\pi}{6}$
$cis\frac{2(4)\pi}{6} = -cis\frac{2(4 - 3)\pi}{6} = -cis\frac{2(1)\pi}{6}$
$cis\frac{2(5)\pi}{6} = -cis\frac{2(5 - 3)\pi}{6} = -cis\frac{2(2)\pi}{6}$

For the last round of pure math teaching period, I’d come across other questions from F.7 students, just click here for your further reading.

I still want to say something more including a funny geometric application of “roots of unity" and replies to Justin. Just wait and see.