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Complex triangle equality & Sin(ntheta)/Sin(theta)

  1. Feb 18, 2013 #1
    1. The problem statement, all variables and given/known data
    The first problem is as follows
    2ir7r0m.jpg
    And the second one is
    35hiq0j.jpg


    2. Relevant equations
    For the first question, I do not really know what equations to use. I tried writing it out in the form Z1 = a + ib, Z2 = c + id, etc, but that got me nowhere.

    The second question has been troublesome for me too; I've written it out in exponential form, (e-in[itex]\theta[/itex]-ein[itex]\theta[/itex])/(e-i[itex]\theta[/itex]-ei[itex]\theta[/itex])

    but I do not see how anything will follow from that, either.

    3. The attempt at a solution

    For the first question, my intuition tells me (well, it's probably just really obvious) that an equilateral triangle satisfies the equation. Showing this is.. fairly straightforward I guess, as they are just all equal to each other. However, I doubt that this is the entire question, there is more to it. Could you help me in the right direction?

    For the second question, I am even more clueless. You have to understand that this is the 2nd week of my course in complex analysis, so we have covered hardly anything; we've only just done the elementary functions, some limits and continuity, nothing fancy at all. My first intuition was de moivre's theorem, but that doesn't apply to this either I guess. Again, if you could provide me with a simple hint, that would be amazing.

    Kind regards
     
  2. jcsd
  3. Feb 18, 2013 #2

    jbunniii

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    I'm not sure if this will help, but the left hand side of the second equation can be rewritten in terms of complex exponentials. If we start with the following identity
    $$\sum_{k=0}^{n-1}r^k = \frac{1-r^n}{1-r}$$
    and put ##r = e^{i2\theta}##, we get
    $$\sum_{k=0}^{n-1}e^{i2\theta k} = \frac{1-e^{i2\theta n}}{1-e^{i2\theta}} = \frac{e^{i\theta n}}{e^{i\theta}} \frac{\sin(\theta n)}{\sin(\theta)} = e^{i \theta(n-1)} \frac{\sin(\theta n)}{\sin(\theta)}$$
    Therefore,
    $$\frac{\sin(\theta n)}{\sin(\theta)} = e^{-i \theta(n-1)} \sum_{k=0}^{n-1}e^{i2\theta k}$$
    which can be simplified further by a change of indices. Maybe try writing the right hand side in terms of complex exponentials as well?
     
  4. Feb 18, 2013 #3

    jbunniii

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    For the first part, put ##w_1 = z_1 - z_2##, ##w_2 = z_2 - z_3##, ##w_3 = z_3 - z_1##. Then we have ##w_1 + w_2 + w_3 = 0## and ##w_1^2 + w_2^2 + w_3^2 = 0##. Therefore,
    $$0 = (w_1 + w_2 + w_3)^2 = w_1^2 + w_2^2 + w_3^2 + 2w_1 w_2 + 2w_2 w_3 + 2w_3 w_1$$
    which means that
    $$w_1 w_2 + w_2 w_3 + w_3 w_1 = 0$$
    This is of course satisfied if $$w_1 = w_2 = w_3 = 0$$
    and it's also clear that if any of ##w_1##, ##w_2##, or ##w_3## is zero, then they must all be zero. So suppose they are all nonzero. The natural thing to do is to start dividing. See if you can manipulate the resulting expressions to show that ##|w_1| = |w_2| = |w_3|##.
     
  5. Feb 18, 2013 #4
    Hmm, the first hint, about the sinus n theta, does make some sense. I'll try and work on that a bit more. However, the second one, I don't really see how dividing is the natural thing to do. Up to that point your derivation makes a lot of sense (and is very clever indeed), but how would I continue? Write it out in polar coordinates, or just the euclidian version?
     
  6. Feb 18, 2013 #5

    jbunniii

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    No, you don't need to write it out in coordinates at all. Start with
    $$w_1 w_2 + w_2 w_3 + w_3 w_1 = 0$$
    and divide by the product of any pair of the ##w_i##'s. For example, if we divide by ##w_3 w_1##, we get
    $$\frac{w_2}{w_3} + \frac{w_2}{w_1} + 1 = 0$$
    Then combine the second and third terms:
    $$\frac{w_2}{w_3} + \frac{w_2 + w_1}{w_1} = 0$$
    Now, what is ##w_2 + w_1## equal to?
     
  7. Feb 18, 2013 #6
    So w1+w2 = -w3. Ok, from that I get to w1/w2 = w2/w3 = w3/w1
    I apologize for being so stupid about this, I'm just really not used to manipulating these things yet
     
  8. Feb 18, 2013 #7

    jbunniii

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    Good, so now take magnitudes:
    $$\frac{|w_1|}{|w_2|} = \frac{|w_2|}{|w_3|} = \frac{|w_3|}{|w_1|}$$
    How can you use this to conclude that ##|w_1| = |w_2| = |w_3|##?
     
  9. Feb 18, 2013 #8
    Ah man, I feel so ridiculously stupid. It sounds like very basic arithmetic, but I just don't see it.

    Edit: I see it now, thanks a ton, your help has been great!
     
    Last edited: Feb 18, 2013
  10. Feb 18, 2013 #9

    jbunniii

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    One more thing: notice that this shows that any solution must be an equilateral triangle. But it does not show that every equilateral triangle is a solution. You will have to show that separately. Fortunately, this is easier. Hint: the triangle is equilateral if and only if ##w_1##, ##w_2##, and ##w_3## all have the same magnitude, and their phases must be of the form ##e^{i\theta}##, ##e^{i\theta + \pi/3}##, and ##e^{i\theta - \pi/3}##. Show that if these conditions are satisfied, then ##w_1^2 + w_2^2 + w_3^2 = 0##.
     
  11. Feb 19, 2013 #10
    Thanks a lot, I completely forgot about that part.

    Now, for the second question, not much has worked out so far. I've never actually worked with a sequence of multiplications, and I have no clue how to rewrite that in terms of a sum of complex exponentials.. If anyone could provide a hint, that would be great.
     
  12. Feb 19, 2013 #11

    tiny-tim

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    Hi Verdict! :smile:

    (i suspect you're not suppopsed to do it this way :redface:, but …)

    Hint: sinnθ/sinθ is obviously a polynomial in cosθ of degree n-1 :wink:

    (and the polynomial xn - 1 will also come in handy)
     
  13. Feb 19, 2013 #12
    I'm sorry, at the risk of sounding like an idiot, but how is it a polynomial in cosine theta of degree n-1? Writing it out in exponentials did not make it any more clear for me :(

    For xn-1, I do indeed know how to factor that, as its just (x-1)(xn-1+xn-2.....+x+1)
    And in the complex case, zn = 1 are just the roots of unity, I guess.
     
    Last edited: Feb 19, 2013
  14. Feb 19, 2013 #13

    tiny-tim

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    two ways of showing it …

    i] sinnθ = sin(n-1)θcosθ + cos(n-1)θsinθ, and just keep going, and you'll find all the odd powers of sinθ cancel, and the even ones can be converted to (1 - cos2θ)

    ii] write isinnθ = (cosθ + isinθ)n - (cosθ - isinθ)n, divide by sinθ, and again you've only even powers of cosθ and sinθ :wink:

    EDIT: ah, i've just seen your edit …
    yes, and the important thing is that that gives you eg the sum of the nth roots of 1 (= … ?)

    however, you won't need that polynomial until you've found the roots of the polynomial in cosθ
     
    Last edited: Feb 19, 2013
  15. Feb 19, 2013 #14
    Hmm, I tried doing the second version, with de moivre's theorem that would be cos(nθ)+isin(nθ) - cos(nθ) + isin(nθ), wouldn't that be equal to 2 times isin(nθ)? Or is it simply because the theorem doesn't work for (cosθ - isinθ)n Again, at the risk of being an idiot..
    Thinking of it, it is just equal to e-e-iθ, which is indeed(?) 2isinθ, right?

    The nth roots of 1 are e2∏ik/n, with k going from 0 to n-1, right?
     
    Last edited: Feb 19, 2013
  16. Feb 19, 2013 #15

    tiny-tim

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    yes, isin(nθ) = isin(nθ) ! :smile:

    (I was a factor of 2 out :redface:)

    but you dont want a function of nθ, you want functions of θ only, which is what my formula gives you :smile:
     
  17. Feb 19, 2013 #16
    Alright, so according to ii],
    sin(nθ)/sin(θ) = ((cosθ + isinθ)n - (cosθ - isinθ)n) / 2isinθ

    Which is of course simpler in exponentials. I'm sorry if I'm heading in the wrong direction! (disregard the removed, stupid question)
     
    Last edited: Feb 19, 2013
  18. Feb 19, 2013 #17

    tiny-tim

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    ok :smile:

    now can you see that a lot of terms on the top cancel out?

    what are you left with? :wink:
     
  19. Feb 19, 2013 #18
    Uhm, it sounds like the binomial theorem, so for example (cosθ + isinθ)n has n+1 terms of sine, multiplied by a cosine, and some factor.. I'm sorry, this is quite terrible, but how does that make so many of them drop out? Aren't you dividing higher order sines by first order sines?
     
  20. Feb 19, 2013 #19

    tiny-tim

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    (cosθ + isinθ)n - (cosθ - isinθ)n

    isn't it obvious that all the terms in the left half are either equal to or are minus the terms in the right half? :redface:
     
  21. Feb 19, 2013 #20
    Alright yeah, I suppose so indeed, so a big part cancels by itself. Quantifying which part that is exactly though..
     
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