# Quod Erat Demonstrandum

## 2007/12/18

### [AL][PM] Applications of roots of unity

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 11:42 上午
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Oh, it’s the time for switching the channel into English. Actually, I’m afraid to write in English because I’m blamed not to use English to write blog too often, it may do harm to students. (My English is poor ma, right!) But, firstly, there are less than 10 students reading this blog, so it may be harmless relatively. And next, because of the beautiful fonts, I try to use English in this blog, and that is the only reason ^_^!

F.7C students were just given a quiz, here are some old stuff for your (so-called) enrichment.

Q.1 [Remainder theorem?]

Show that $x^{1987} + x^{1997} + x^{2007}$ is divisible by $x^{1987} + x^{1988} + x^{1989}$.

Q.2 [Binomial theorem?]

Evaluate $C_{0}^{2007} + C_{4}^{2007} + C_{8}^{2007} + \dots + C_{2004}^{2007}$.

Q.3 [Geometry problem?]

Let $P_1, P_2, \dots P_{2007}$ be the vertices of a regular 2007-gon inscribed in a unit circle. Evaluate $P_1P_2 \times P_1P_3 \times \dots \times P_1P_{2007}$.

The questions above may look different to each other, however, by using the roots of unity, we can solve them.

Solution to Q.1

Let $\omega$ be a complex cube root of unity, i.e. $\omega^3 = 1$ and $\omega \neq 1$.

If $P(x)$ is a polynomial (over $\mathbb{R}$) such that $P(\omega) = P(\omega^2) = 0$, then $(x - \omega)(x - \omega^2)$ is a factor of $P(x)$. Hence $P(x)$ is divisible by $1 + x + x^2$.

Now, let $P(x) = 1 + x^{10} + x^{20}$

It is easy to check

$1 + \omega^{10} + \omega^{20} = 1 + \omega^{20} + \omega^{40} = 0$ (Why?)

Hence $P(\omega) = P(\omega^2) = 0$; thus $P(x)$ is divisible by $1 + x + x^2$. Or

$1 + x^{10} + x^{20} = (1 + x + x^2)Q(x)$ for some polynomial $Q(x)$

Now, $x^{1987} + x^{1997} + x^{2007}$
= $x^{1987}(1 + x^{10} + x^{20})$
= $x^{1987}(1 + x + x^2)Q(x)$
= $(x^{1987} + x^{1988} + x^{1989})Q(x)$

Result follows.

Solution to Q.2

Let $\omega$ be a complex forth root of unity, i.e. $\omega^4 = 1$ and $\omega \neq 1$.

The following results are trivial.

$1 + \omega^k + \omega^{2k} + \omega^{3k} = 4$ if k is a multiple of 4
$1 + \omega^k + \omega^{2k} + \omega^{3k} = 0$ if k is not a multiple of 4

Now

$(1 + x)^{n} = C_{0}^{n} + C_{1}^{n}x + C_{2}^{n}x^2 + \dots + C_{n}^{n}x^n$ (where n = 2007)

Put $x = 1, \omega, \omega^2, \omega^3$ into the above accordingly. We have

$2^{n} = C_{0}^{n} + C_{1}^{n} + C_{2}^{n} + \dots + C_{n}^{n}$
$(1 + \omega)^{n} = C_{0}^{n} + C_{1}^{n}\omega + C_{2}^{n}\omega^2 + \dots + C_{n}^{n}\omega^n$
$(1 + \omega^2)^{n} = C_{0}^{n} + C_{1}^{n}\omega^2 + C_{2}^{n}\omega^4 + \dots + C_{n}^{n}\omega^{2n}$
$(1 + \omega^3)^{n} = C_{0}^{n} + C_{1}^{n}\omega^3 + C_{2}^{n}\omega^6 + \dots + C_{n}^{n}\omega^{3n}$

Sum up the above 4 equations and use the trivial results just mentioned, we have

$4(C_{0}^{2007} + C_{4}^{2007} + C_{8}^{2007} + \dots + C_{2004}^{2007})$
= $2^{2007} + (1 + \omega)^{2007} + (1 + \omega^2)^{2007} + (1 + \omega^3)^{2007}$

Hey, we can simplify it further because we may take $\omega = i, \omega^2 = -1, \omega^3 = -i$, hence

$C_{0}^{2007} + C_{4}^{2007} + C_{8}^{2007} + \dots + C_{2004}^{2007}$
= $\frac{1}{4}(2^{2007} + (1 + i)^{2007} + (1 - 1)^{2007} + (1 - i)^{2007})$
= $\frac{1}{4}(2^{2007} + 2^{\frac{2007}{2}}(2\cos\frac{2007\pi}{4}))$
= $\frac{1}{4}(2^{2007} + 2^{\frac{2007}{2}}(2\cos\frac{\pi}{4}))$
= $2^{1002}(2^{1003} + 1)$

Solution to Q.3

The complex numbers corresponding to the vertices of a regular n-gon inscribed in a unit circle are roots of $z^n = 1$. Hence, we may let

$1, \omega, \omega^2, \dots, \omega^{2006}$ be the complex numbers corresponding to the vertices of the regular 2007-gon; where $\omega = \cos\frac{2\pi}{2007} + i\sin\frac{2\pi}{2007}$.

It is easy to have $\omega, \omega^2, \dots, \omega^{2006}$ are roots of the equation

$x^{2006} + x^{2005} + \dots + x + 1 = 0$ – – – (1)

Suppose we translate the regular 2006-gon to the left by 1 unit, then the vertices will become $0, \omega - 1, \omega^2 - 1, \dots, \omega^{2006} - 1$, and hence

$P_1P_2 = |\omega - 1| = |z_1| (say)$
$P_1P_3 = |\omega^2 - 1| = |z_2|$
$P_1P_4 = |\omega^3 - 1| = |z_3|$

$P_1P_{2007} = |\omega^{2006} - 1| = |z_{2006}|$

Now, if we can find out an equation whose roots are $z_1, z_2, \dots , z_{2006}$, then $P_1P_2 \times P_1P_3 \times \dots \times P_1P_{2006}$ is simply the modulus of the product of roots.

By (1), just set $z = x - 1$ then an equation with roots $z_1, z_2, \dots , z_{2006}$ is

$(z + 1)^{2006} + (z + 1)^{2005} + \dots + (z + 1) + 1 = 0$
$z^{2006} + \dots + 2007 = 0$

Since the product of roots of the above equation = 2007,
$P_1P_2 \times P_1P_3 \times \dots \times P_1P_{2006} = 2007$

Not enough? If you want to know how powerful in using complex numbers for solving elementary mathematics problems, I suggest an old Chinese popular mathematics for your own leisure reading: 神奇的複數─如何利用複數解中學數學難題

I do admire Mr. Siu’s concept : “I don’t teach: I share". Read his blog to learn better English!
http://hk.myblog.yahoo.com/siu82english

## 6 則迴響 »

1. Ar… it seems to be a long time since I visit your blog.
During exam, I’ve got no time, and checking my email, I think I’m gonna spend some time reading them.
God! You still got momentum writing blog during exam… though I notice it is not your exam, …"
Oh, tmr xmas party >< the previous Xmas party in SFX, is… is ‘oh my ears’. you know?… hehe

迴響 由 Ed' — 2007/12/20 @ 7:31 下午 | 回應

2. Oh, it has been quite some while I haven’t checked this blog. During exam, I just don’t wanna spend time on blog’ing’.
I can’t figure out in what way you can find momentum to write blog while exam,… though it’s not yours, I notice.
Oh, tmr, Xmas party… The previous party in SFX, it is… is…’oh my ears’
Well, your busy time come, checking papers, hehe

迴響 由 Ed' — 2007/12/20 @ 7:57 下午 | 回應

3. woops, sorry I post twice=.=

迴響 由 Ed' — 2007/12/21 @ 1:11 下午 | 回應

4. I was not that busy during the exam period, and therefore I’d spent time on blog writing. Wish you and all Xaverians a merry Christmas and a happy New Year! Come on 2008!

迴響 由 johnmayhk — 2007/12/23 @ 2:47 下午 | 回應

5. HAPPY NEW YEAR! (though it is late…)

迴響 由 Ed' — 2008/01/02 @ 5:47 下午 | 回應

6. Thanks for these examples. They are very useful for us studying AL Pure Maths.

迴響 由 杜輝 — 2008/11/22 @ 7:00 下午 | 回應