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If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $$\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$$

Here's how I tried it

$$x+y+z=xyz $$

So, by De Moivre's Theorem, $$(\cos a + \cos b + \cos c) + i(\sin a + \sin b + \sin c) = \cos(a+b+c) + i \sin(a+b+c) $$

Equating real and imaginary parts, $$\cos a + \cos b + \cos c= \cos(a+b+c)$$ and similarly for sine. Now,

$$(a-b) + (b-c) + (c-a) =0$$

What to do now? Please help. And please use De Moivre Theorem!

Blue
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Pratim Das
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  • You could expand $\cos(a+b+c)$ and $\sin(a+b+c)$. – amsmath Aug 31 '18 at 02:08
  • Just point out where I am wrong. If for the complex numbers $x$, $y$, $z$; $x+y+z=xyz$, then it can be shown that $\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=1$ (do it in reverse) which easily leads to $\cos (x-y)+\cos (y-z)+\cos (z-x) =1$. What am I missing? –  Aug 31 '18 at 06:03

3 Answers3

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Note that $$\cos(a-b) = \cos(a)\cos(b)+\sin(a)\sin(b) = \Re(x\bar{y})$$ So $$\label{eq1}\cos(a-b)+\cos(b-c)+\cos(c-a) = \Re(x\bar{y}+y\bar{z}+z\bar{x})$$ With that in mind, note that just like $x$, $y$ and $z$, the $xyz$ complex number is of modulus 1. So if we go back to the $x+y+z=xyz$ identity, we see that $$|x+y+z|=1$$ In other words, $$\begin{split} 1 & = |x+y+z|^2 \\ & = (x+y+z)(\bar x+\bar y +\bar z) \\ & = x\bar x + x \bar y + x \bar z + y\bar x + y \bar y + y \bar z + z \bar x + z \bar y + z \bar z \\ & = 1 + x \bar y + x \bar z + 1 + y\bar x + y \bar z + 1 + z \bar x + z \bar y \\ & = 3 + (x \bar y + y \bar z + z \bar x) + (y\bar x + z \bar y + x \bar z) \\ & = 3 + 2 \Re(x \bar y + y \bar z + z \bar x) \end{split}$$ We conclude that $$\Re(x\bar{y}+y\bar{z}+z\bar{x})+1=0$$ which yields the desired result.

Stefan Lafon
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(This is a followup on Stefan Lafon's answer, too long for a comment.)

Using that $\,|x|=1 \iff \bar x = \dfrac{1}{x}\,$ and $\,\cos(a-b)= \operatorname{Re}\left(\dfrac{x}{y}\right)=\operatorname{Re}(x \bar y) = \operatorname{Re}(\bar x y)\,$, it follows that:

$$ \begin{align} \cos(a-b) + \cos(b-c) + \cos(c-a) &= \cos(a-b) + \cos(a-c) + \cos(b-c) \\ &= \operatorname{Re}\left(x\bar y+ x \bar z + y \bar z\right) \\ &= \operatorname{Re}\left(x\left(\bar y + \bar z\right) + y \bar z\right) \\ &= \operatorname{Re}\left(x\left(\bar x \bar y \bar z - \bar x\right) + y \bar z\right) \\ &= \operatorname{Re}\left(\bar y \bar z + y\bar z -1 \right) \\ &= \operatorname{Re}\left(\left(y + \bar y\right) \bar z -1 \right) \\ &= -1 + 2 \operatorname{Re}(y) \operatorname{Re}(z) \end{align} $$

For the latter RHS to be $\,-1\,$, the second term must be zero, so one of $\,y,z\,$ must be a purely imaginary number of modulus $\,1\,$ i.e. $\,\pm i\,$. With some more legwork, it follows that the solution set of the given constraints is $\,\{(x,-x,z)\mid x=\pm i, |z| = 1\}\,$ or permutations thereof.

dxiv
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Squaring & adding we get

$$\cos^2(a+b+c)+\sin^2(a+b+c)=(\cos a+\cos b+\cos c)^2+(\sin a+\sin b+\sin c)^2$$

$$\implies1=1+1+1+2\sum_{\text{cyc}(a,b,c)}\cos(a-b)$$

Observe that we actually don't need $x+y+z=xyz$

The sufficient condition is $$ x+y+z=\cos p+i\sin p\iff|x+y+z|=1$$

lab bhattacharjee
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