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I'm struggling to figure out this question. Part a) and b) I can do fine, but part c) has me stumped.

Question: a) Find the ninth roots of unity

b) Hence show that: $$z^6+z^3+1=(z^2-2\cos\frac{2\pi}9+1)(z^2-2\cos\frac{4\pi}9+1)(z^2-2\cos\frac{8\pi}9+1)$$ c)Deduce that: $$2\cos(3\theta)+1=8\left(\cos\theta-\cos\frac{2\pi}9\right)\left(\cos\theta-\cos\frac{4\pi}9\right)\left(\cos\theta-\cos\frac{8\pi}9\right)$$

Jyrki Lahtonen
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spuddy
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  • Is there actually a connection to modular forms? I'm afraid I fail to see it, but I'm not an expert on them. Looks like you only need basic facts about writing trig functions with complex exponentials. – Jyrki Lahtonen Oct 26 '19 at 05:47

2 Answers2

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Divide both sides by $z^3$

$$z^3+\dfrac1{z^3}+1=\prod_{r=1}^3\left(z+\dfrac1z-2\cos\dfrac{2^r\pi}9\right)$$

Now set $z=e^{i\theta}$

Use Intuition behind euler's formula to find

$z^m+\dfrac1{z^m}=2\cos m\theta$ for integer $m$

lab bhattacharjee
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Well then $9$ ninth roots of unity are $r_k = \cos \frac {2k\pi}9 + i \sin\frac {2k\pi}9$ for $k = 0..... 9$.

Now $(z^6 + z^3 + 1)(z^3 - 1) = z^9 -1 = (z-r_0)(z-r_1)(z-r_2)....(z-r_8)$.

And $(z^6 + z^3 + 1) = \frac {(z-r_0)(z-r_1)(z-r_2)....(z-r_8)}{z^3 -1}$.

Now notice the $3$ third roots of unity are $s_j = \cos \frac {2j\pi}3 + i\sin \frac {2j\pi}3= \cos \frac {2*3j\pi}9 + i\sin \frac {2*3j\pi}9 = r_{3j}$.

So the three third roots of unity or $r_0, r_3, $ and $r_6$ and $z^3-1 = (z-r_0)(z-r_3)(z-r_6)$.

So $(z^6 + z^3 + 1) = \frac {(z-r_0)(z-r_1)(z-r_2)....(z-r_8)}{z^3 -1}=$

$\frac {(z-r_0)(z-r_1)(z-r_2)....(z-r_8)}{(z-r_0)(z-r_3)(z-r_6)} = $

$(z-r_1)(z-r_2)(z-r_4)(z-r_5)(z-r_7)(z-r_8)=$

$[(z-r_1)(z-r_8)][(z-r_2)(z-r_7)][(z-r_4)(z-r_5)]$.

Now notice that $[(z-r_k)(z-r_{9-k})] =$

$(z - (\cos \frac {2k\pi}9 + i\sin\frac {2k\pi}9))(z - (\cos \frac {2(9-k)\pi}9 + i\sin\frac {2(9-k)\pi}))=$.

(remember your basic trig $\cos|\sin (2\pi + x) = \cos|\sin x$ and $\cos(-x)= \cos x$ and $\sin (-x) = -\sin x$. So....)

$\cos \frac{2(9-k)\pi}9 = \cos \frac {18\pi -2k\pi}=\cos\frac {-2k\pi}9 = \cos \frac {2k\pi}9$

And $\sin\frac{2(9-k)\pi}9 = \sin \frac {18\pi -2k\pi}=\sin\frac {-2k\pi}9 = -\sin \frac {2k\pi}9$

So

$[(z-r_k)(z-r_{9-k})] =$

$(z - (\cos \frac {2k\pi}9 + i\sin\frac {2k\pi}9))(z - (\cos \frac {2(9-k)\pi}9 + i\sin\frac {2(9-k)\pi}))=$

$(z - [\cos \frac {2k\pi}9 + i\sin\frac {2k\pi}9])((z - [\cos \frac {2k\pi}9 - i\sin\frac {2k\pi}9])=$

$z^2 -z([\cos \frac {2k\pi}9 + i\sin\frac {2k\pi}9]+ [\cos \frac {2k\pi}9 - i\sin\frac {2k\pi}9]) + [\cos \frac {2k\pi}9 + i\sin\frac {2k\pi}9][\cos \frac {2k\pi}9 - i\sin\frac {2k\pi}9]=$

$z^2- 2\cos\frac {2k\pi}9 + (\cos^2 \frac {2k\pi}9 + \sin^2\frac {2k\pi}9)=$

$z^2 - 2\cos\frac {2k\pi}9 + 1$.

And that's it....

$z^6 +z^3 + 1 = [(z-r_1)(z-r_8)][(z-r_2)(z-r_7)][(z-r_4)(z-r_5)]=$

$(z^2 -2\cos\frac {2\pi}9 + 1)(z^2- 2\cos \frac {4\pi}9+1) (z^2 - 2\cos\frac {6\pi}9+1)$.

fleablood
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