Complex Roots

**bobak** · March 20th 2008, 09:03 AM

the first part is straightforward. for the roots i get $\alpha^0 , \alpha^1 , \alpha^2 , \alpha^3 , \alpha^4 , \alpha^5 , \alpha^6$ .

For the quadratic I guessed got the root as the complex conjugate which is $\alpha^{-1} + \alpha^{-2} \alpha^{-4}$

i get that $A= - 2 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7} \right)$

and B is the product of the roots, but I am having trouble showing the required results.

**ThePerfectHacker** · March 20th 2008, 09:55 AM

If $\alpha + \alpha^2 + \alpha^4$ is a root then $\bar \alpha + (\bar \alpha)^2 + (\bar \alpha)^4$ , i.e. conjugate is a root. Note, $\alpha \bar \alpha = |\alpha|^2 \implies \alpha \bar \alpha = 1$ , thus, $\bar \alpha = \alpha^{-1}$ . But since $\alpha \cdot \alpha^6 = 1 \implies \alpha^{-1} = \alpha^6$ . Thus, $\bar \alpha = \alpha^6$ . This immediately implies, $( \bar \alpha)^2 = \alpha^5$ and $( \bar \alpha)^4 = \alpha^3$ . Thus, $\alpha^3 + \alpha^5 + \alpha^6$ is the other root.

This means (by Viete) that $A = - (\alpha + \alpha^2 + \alpha^4 + \alpha^3+\alpha^5+\alpha^6) = 1$ . Because we are using the fact that the sum of the roots of unity is zero, i.e. $1+\alpha+...+\alpha^6 = 0$ .

Now the product is, $(\alpha + \alpha^2 + \alpha^4)(\alpha^3 + \alpha^5 + \alpha^6) = (\alpha + \alpha^2 + ... + \alpha^7) + 2\alpha^7 = 2$ . Here we would the fact that $\alpha^8 = \alpha,\alpha^9 = \alpha^2,\alpha^{10} = \alpha^3$ . Thus, we have that $B=2$ .

This means that: $z_1 = \alpha+\alpha^2+\alpha^4$ and $z_2 = \alpha^3+\alpha^5+\alpha^6$ solve $z^2 + z + 2 = 0$ .

Now, $-\frac{1}{2}+ i \frac{\sqrt{7}}{2}, - \frac{1}{2}-i\frac{\sqrt{7}}{2}$ are its solutions. Thus, they are actually $z_1,z_2$ (not necessarily in that order).

This means, $\sin \frac{2\pi}{7}+\sin \frac{4\pi}{7}+\sin\frac{8\pi}{7} = \pm \frac{\sqrt{7}}{2}$ .
You need to determine the sign. That is what the problem says.

**bobak** · March 20th 2008, 10:45 AM

I followed that really well, Thank you TPH.

I get the sign as positive. as $\sin\frac{8\pi}{7} = - \sin\frac{\pi}{7}$

and $\sin\frac{2\pi}{7} > \sin\frac{\pi}{7}$

$\Rightarrow \sin\frac{2\pi}{7} - \sin\frac{\pi}{7} > 0$

$\sin\frac{4\pi}{7} > 0$ as it is second quadrant

therefore $\sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} - \sin\frac{\pi}{7} > 0$

that good enough?

**ThePerfectHacker** · March 20th 2008, 12:04 PM

Originally Posted by bobak

that good enough?

Looks good.

**bobak** · March 22nd 2008, 06:40 PM

I am just modifying this question of prove your current signature TPH.

we have $\sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7} = \frac{\sqrt{7}}{2}$

$\Rightarrow 2 \sin\frac{2\pi}{7} + 2 \sin\frac{4\pi}{7} + 2 \sin\frac{8\pi}{7} = \sqrt{7}$

we will use $\sin\frac{2\pi}{7} = \sin\frac{5\pi}{7}$
$\sin\frac{4\pi}{7} = \sin\frac{3\pi}{7}$
and $\sin\frac{8\pi}{7} = -\sin\frac{\pi}{7} = -\sin\frac{6\pi}{7}$

using all of that together we have proved your signature

$-\sin \frac{\pi}{7} + \sin \frac{2\pi}{7} + \sin \frac{3\pi}{7} + \sin \frac{4\pi}{7}+\sin \frac{5\pi}{7} - \sin \frac{6\pi}{7} = \sqrt{7}$

**ThePerfectHacker** · March 22nd 2008, 06:56 PM

How about this?

$\sin \frac{\pi}{11} + \sin\frac{2\pi}{11}+\sin \frac{3\pi}{11} -\sin \frac{4\pi}{11}+\sin\frac{5\pi}{11}+\sin\frac{6\pi }{11}-\sin \frac{7\pi}{11}+\sin\frac{8\pi}{11}+\sin\frac{9\pi }{11}+\sin\frac{10\pi}{11}$ $=\sqrt{11}$ .

How about this?
$\sin \frac{\pi}{11}\sin \frac{2\pi}{11}\sin \frac{3\pi}{11}\sin\frac{4\pi}{11}\sin\frac{5\pi}{ 11}\sin \frac{6\pi}{11}\sin\frac{7\pi}{11}\sin \frac{8\pi}{11}\sin \frac{9\pi}{11}\sin \frac{10\pi}{11} = \frac{11}{2^{10}}$ .

By the way. Identity #2 generalizes.

**bobak** · March 22nd 2008, 07:03 PM

the Gauss Sum stuff your referring to looks like it is beyond me (for now).

by generalises do you mean $\prod_{k=1}^{n-1}{\sin \left( \frac{k\pi }{n} \right)\; =\; \frac{n}{2^{n-1}}}$

**ThePerfectHacker** · March 22nd 2008, 07:16 PM

Originally Posted by bobak

by generalises do you mean $\prod_{k=1}^{n-1}{\sin \left( \frac{k\pi }{n} \right)\; =\; \frac{n}{2^{n-1}}}$

Exactly. It happens to be not hard at all to derive (but still impressive). If you are familar with roots of unity. Here.

Thread: Complex Roots

LinkBack

Thread Tools

Display

Complex Roots

Similar Math Help Forum Discussions

Complex 5th Roots

complex roots

Complex Roots

Complex roots

complex roots

Search Tags