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Let $z_j=r(cosφ_j+isinφ_j), r\in R$ for $j=1,2,3$ be different complex numbers.

If the numbers $w_1=z_1+z_2z_3$, $w_2=z_2+z_1z_3$, $w_3=z_3+z_1z_2$ are real, prove that $z_1z_2z_3=1$

I know one solution for this problem, but it just seems too complex. Wonder if there is anything simpler. (i won't post the solution that i know, at least for now)

Gerry Myerson
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  • Your first sentence seems superfluous aside from the fact that the $z_j$ are complex. (This polar form may come in handy, to be sure, but it's not essential to the statement of the problem...) – Semiclassical Sep 23 '14 at 21:51
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    it tells you that they have the same modulus –  Sep 23 '14 at 21:56
  • You're quite correct, I was being careless. Though if they have the same modulus, then their product being one means it's necessary (though not sufficient) for that modulus to be $1$. – Semiclassical Sep 23 '14 at 21:56
  • Actually, it would be helpful if you'd indicate briefly the method you already possess. Otherwise we run the risk of telling you something you already know. (Plus you can post it as an answer and get feedback/comments/rep.) – Semiclassical Sep 23 '14 at 22:13
  • This particular method proves that $sin(φ_1+φ_2+φ_3)=0$ with contradiction. –  Sep 23 '14 at 22:22
  • I added the VERY important information that $z_j$ are different. –  Sep 23 '14 at 22:31

1 Answers1

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Hopefully this is not the "too complex" method you were talking about...

$$ z_j = r(\cos \varphi_j + i \sin \varphi_j) = re^{i\varphi_j}, j=1,2,3 $$

Using the parallelogram rule for addition of complex numbers, $0 = \arg(w_1) = \arg(z_1 + z_2 z_3)=\frac1{2} (\arg z_1 + \arg z_2 z_3) = \frac1{2}(\arg z_1z_2z_3) $. This implies that $z_1 z_2 z_3 $ is a positive real number.

$$ z_1 z_2 z_3 = r^3e^{i(\varphi_1 + \varphi_2 + \varphi_3)}>0. \\ \therefore \varphi_1 + \varphi_2 + \varphi_3 = 2n\pi, n \in \mathbb{Z} \\ \because \left|e^{ix}\right| =1, \therefore z_1z_2z_3 = r^3. \\ w_1 = z_1 + \frac{r^3}{z_1}, w_2=z_2 + \frac{r^3}{z_2}, w_3=z_3+\frac{r^3}{z_3}\\ w_j = z_j + r^3z_j^{-1}=r(\cos\varphi_j + i \sin\varphi_j)+r^2(\cos \varphi_j - i \sin \varphi_j)=(r+r^2)\cos \varphi_j + (r-r^2)i \sin \varphi_j, j=1,2,3\\ \Im(w_j)=0\Rightarrow r-r^2 = 0\Rightarrow r=0,1\Rightarrow z_1z_2z_3=0,1. $$ $\Im(z)$ is the imaginary part of $z$. r can't be zero because the precondition stated that the complex numbers were different values - if it were they would be the same i.e. $z_1 = z_2 = z_3 = 0$.

$$ \therefore z_1z_2z_3 = 1 $$

James Harrison
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  • Well, this is not the method I was talking about but it's even more complex for me because I can't even understand it :P –  Sep 25 '14 at 08:13
  • Could you clarify which parts of it are you having trouble understanding? – James Harrison Sep 25 '14 at 08:23
  • I am not familiar with the arg(x) function and also I don't understand those symbols with the three dots you're using. –  Sep 25 '14 at 09:57
  • Here is an explanation of the argument function. Also, $\therefore$ means "therefore" and $\because$ means "because". – James Harrison Sep 25 '14 at 12:25
  • I don't think the argument of the sum is the average of the arguments unless you know that the two numbers have the same modulus. That is, unless you already know that $r=1$. – Jyrki Lahtonen Dec 29 '20 at 23:16
  • Why would the modulus change it? A diagonal through a parallelogram always bisects the angles it touches - that would naturally include the rhombus case you speak of. – James Harrison Jan 22 '21 at 22:49