# Solving two equations

1. Oct 8, 2015

### Frank Einstein

1. The problem statement, all variables and given/known data

Hi everybody, I am trying to find the solution to this system of equations for a bigger physics problem:

h+k+l=2n+1 with n=0,1,2,3...
ei(π/2)(h+k+l)+ei(π/2)(3h+3k+l)+ei(π/2)(3h+k+3l)+ei(π/2)(h+3k+3l)=0

2. Relevant equations

None

3. The attempt at a solution

Ihave tried to solve it by taking the common factor ei(π/2) out and trying to solve
e(h+k+l)+e(3h+3k+l)+e(3h+k+3l)+e(h+3k+3l)=0 by using h+k+l=0; then, I arrive to 1+e2k+2h+e2h+2l+e2k+2l=0.

From here I don't know how to keep going, so if anybody could point me what to do next it wolud be very helpfull.

Last edited by a moderator: Oct 8, 2015
2. Oct 8, 2015

### haruspex

That is a factor of ei(π/2)+x, not of ei(π/2)x.

3. Oct 8, 2015

### RUber

To make these equal to zero, you need to enforce conditions on the periodic functions--both real and imaginary.
$\cos( \frac{ \pi(h+k+l)}{2} ) + \cos( \frac{ \pi(3h+3k+l)}{2} )+\cos( \frac{ \pi(3h+k+3l)}{2} )+\cos( \frac{ \pi(h+3k+3l)}{2} ) =0$
and
$\sin( \frac{ \pi(h+k+l)}{2} ) + \sin( \frac{ \pi(3h+3k+l)}{2} )+\sin( \frac{ \pi(3h+k+3l)}{2} )+\sin( \frac{ \pi(h+3k+3l)}{2} ) =0.$

From these, you should be able to say something about the cases where this must be true.

4. Oct 8, 2015

### RUber

Think of these as :
$\cos( \frac{ \pi(h+k+l)}{2} ) + \cos( \frac{ \pi(h+k+l)}{2} +(h+k)\pi )+\cos( \frac{ \pi(h+k+l)}{2}+ (h+l)\pi )+\cos( \frac{ \pi(h+k+l)}{2}+(k+l)\pi ) =0$
and
$\sin( \frac{ \pi(h+k+l)}{2} ) + \sin( \frac{ \pi(h+k+l)}{2} +(h+k)\pi )+\sin( \frac{ \pi(h+k+l)}{2}+ (h+l)\pi )+\sin( \frac{ \pi(h+k+l)}{2}+(k+l)\pi ) =0.$

5. Oct 8, 2015

### Ray Vickson

Please do not use all bold fonts for your solution; it looks like you are yelling at us!
Mod note: Removed excess bold font

Anyway, if you let $m = 2n + 1$, and if you note that $h+k+l = m$ implies
$$\frac{i \pi}{2} (3h+3k+l) = \frac{i \pi}{2} (3h + 3k + 3l - 2l) = \frac{i \pi}{2} 3m - i \pi l,$$
and so forth, you get a much simpler problem, especially if you multiply through by $e^{-3 m \pi i/2}$.

I would suggest you look at some of the initial cases $m = 1, 3, 5, \ldots$ separately, until you have gained sufficient insight into the nature of the solution for general odd $m$.

Last edited by a moderator: Oct 8, 2015
6. Oct 11, 2015

### Frank Einstein

Thank you very much for your anwsers, guess I will be able to proper solve it.