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

[Verdict]

Homework Statement

The first problem is as follows

And the second one is

Homework 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$\theta$-e^in$\theta$)/(e^-i$\theta$-e^i$\theta$)

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

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

 

[jbunniii]

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?

 

[jbunniii]

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 $|w_1| = |w_2| = |w_3|$.

 

[Verdict]

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?

 

[jbunniii]

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?

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?

 

[Verdict]

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

 

[jbunniii]

So w1+w2 = -w3. Ok, from that I get to w1/w2 = w2/w3 = w3/w1

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 $|w_1| = |w_2| = |w_3|$?

 

[Verdict]

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!

 

[jbunniii]

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 $w_1^2 + w_2^2 + w_3^2 = 0$.

 

[Verdict]

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.

 

[tiny-tim]

Hi Verdict!

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

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

(and the polynomial xⁿ - 1 will also come in handy)

 

[Verdict]

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)(x^n-1+x^n-2...+x+1)
And in the complex case, zⁿ = 1 are just the roots of unity, I guess.

 

[tiny-tim]

… how is it a polynomial in cosine theta of degree n-1?

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

EDIT: ah, I've just seen your edit …

For xⁿ-1, I do indeed know how to factor that, as its just (x-1)(x^n-1+x^n-2...+x+1)
And in the complex case, zⁿ = 1 are just the roots of unity, I guess.

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

 

[Verdict]

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^iθ-e^-iθ, which is indeed(?) 2isinθ, right?

The nth roots of 1 are e_2∏ik/n, with k going from 0 to n-1, right?

 

[tiny-tim]

… that would be cos(nθ)+isin(nθ) - cos(nθ) + isin(nθ), wouldn't that be equal to 2 times isin(nθ)?

yes, isin(nθ) = isin(nθ) !

(I was a factor of 2 out )

but you dont want a function of nθ, you want functions of θ only, which is what my formula gives you

 

[Verdict]

yes, isin(nθ) = isin(nθ) !

(I was a factor of 2 out )

but you dont want a function of nθ, you want functions of θ only, which is what my formula gives you

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)

 

[tiny-tim]

Alright, so according to ii],
sin(nθ)/sin(θ) = ((cosθ + isinθ)ⁿ - (cosθ - isinθ)ⁿ) / 2isinθ

ok

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

what are you left with? ​

 

[Verdict]

ok

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

what are you left with? ​

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?

 

[tiny-tim]

(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?

 

[Verdict]

(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?

Alright yeah, I suppose so indeed, so a big part cancels by itself. Quantifying which part that is exactly though..

 

[tiny-tim]

you don't need to quantify anything

just say which powers of sinθ (odd or even?) disappear, and which powers of sinθ are left

 

[Verdict]

Alright, so by writing it out for n = 3, which is a lousy way I suppose, it seems that the odd powers of sine are left.

 

[tiny-tim]

yup, sinθ, sin³θ, sin⁵θ, etc

now divide by sinθ and then replace every sin²θ by 1 - cos²θ

that gives you an expression with no powers of sinθ,

ie a polynomial in cosθ

i] what degree is it?

ii] how many roots does it have?

iii] what can you say about those roots?​

 

[Verdict]

Uhm, alright. Only the odd sines were left, and these are all reduced by 1 order, so it becomes a polynomial of degree n-1, in sines. Replacing the sines by the cosines in the fashion you suggested, the degree doesn't change. So we get a polynomial in cos, with degree n-1. And a polynomial of degree n-1 also has n-1 roots
Something with the roots of unity?

I have to admit, it is only now that I am realizing what the right hand of the identity I want to prove means.

 

[tiny-tim]

So we get a polynomial in cos, with degree n-1. And a polynomial of degree n-1 also has n-1 roots, but for the wrong reasons I guess that it's actually n-2? Now what can I say about them.

yes, a polynomial in cos, with degree n-1, therefore with a maximum of n-1 different roots

but this polynomial in cosθ is equal to sinnθ/sinθ (remember, that's how we created it!) …

so can you think of n-1 different values of θ for which sinnθ/sinθ = 0 ? ​

 

[Verdict]

Yes, that would be.. uhm, θ = pi k / n, but how to limit it to n-1 values.. k = 1 to k = n-1? k = 0 won't work because of the denominator, and k = n doesn't work for the same reason.

 

[tiny-tim]

ok, now look at the original question, and think!

 

[Verdict]

Alright, so on the right hand side, we have a polynomial of cosθ^n-1 + ... etc.
We have found the roots of the left hand side, θ = pi k / n. These are thus also the roots of the right hand side, of the cosines. And a polynomial can be rewritten as a product of its roots, exactly as has been written on the right hand side, apart from the factor of 2^n-1! As in, (cosθ-cos(pi/n)*(cosθ-cos(2pi/n)...x(cosθ-cos((n-1)pi/n))

So basically, we have done that part? The factor of 2^n-1 though..

 

[tiny-tim]

So basically, we have done that part? The factor of 2^n-1 though..

yes, that's exactly right

you have shown that the LHS is equal to the RHS times a factor

now find that factor by trying an easy value of θ, say θ = π/2 ​

 

[Verdict]

Hmm. On the LHS, using theta = pi/2 makes the denominator 1, and the top takes.. Well that depends on n, doesn't it? Can I say without knowing n?

Looking at RHS, the first term just disappears. The second term though, again, can I say much about it without knowing n?

Thanks a ton for the amazing help so far though, I can see that I am not very clever to work with today..

 

[tiny-tim]

Looking at RHS, the first term just disappears. The second term though, again, can I say much about it without knowing n?

try it first (ie θ = π/2) for n odd, then the RHS is … ?

then try it for n even, but first divide both sides by cosθ

 

[Verdict]

Alright, so I tried for n = 5, which gives 1/16 on RHS and 1 on LHS, so indeed the factor works out as 2^(n-1). Not a general proof of course.. Still have to figure that one out. Is there an identity for this?
But for the even, how can you divide by cos theta if you set it is equal to 0? (pi/2 for theta)

 

[tiny-tim]

Alright, so I tried for n = 5, which gives 1/16 on RHS and 1 on LHS, so indeed the factor works out as 2^(n-1). Not a general proof of course.. Still have to figure that one out. Is there an identity for this?

hmm … even for odd n, it's not as easy as i thought it would be

But for the even, how can you divide by cos theta if you set it is equal to 0? (pi/2 for theta)

you can divide both sides by cosθ first, to give you a general equation, and then put cosθ = 0

(and I'm off to bed now :zzz:)

 

[Verdict]

Yeah, I can't really see an easy method too :(
And yes, dividing by cosine first is the only thing that makes sense I suppose, but that makes it a bit more complicated than it was! It was all so pretty..

Anyway, thanks a lot for the help, I think I will give it a rest for now too and look at it again tomorrow.

 

[Verdict]

The professor suggested looking at Chebyshev polynomials, does that make any sense to you?