MHB A complex number problem that has been vexing me....

[MarkFL]

Recently, on another forum, the following problem was posted:

Given three distinct complex numbers:

$\displaystyle z_1,z_2,z_3$

where:

$\displaystyle |z_1|=|z_2|=|z_3|\ne0$

and:

$\displaystyle z_1+z_2z_3,z_2+z_1z_3,z_3+z_1z_2$

are all real, then prove:

$\displaystyle z_1z_2z_3=1$

I began with:

$\displaystyle z_1=re^{\theta_1 i}$

$\displaystyle z_2=re^{\theta_2 i}$

$\displaystyle z_3=re^{\theta_3 i}$

where $\displaystyle 0\le\theta_n<2\pi$

For the 3 expression that are real, I equated the imaginary parts to zero, which gives:

$\displaystyle \sin(\theta_1)+r\sin(\theta_2+\theta_3)=0$

$\displaystyle \sin(\theta_2)+r\sin(\theta_1+\theta_3)=0$

$\displaystyle \sin(\theta_3)+r\sin(\theta_1+\theta_2)=0$

I next solved the first two equations for r, and equated, to find:

$\displaystyle \sin(\theta_1)\sin(\theta_1+\theta_3)=\sin(\theta_2)\sin(\theta_2+\theta_3)$

Using a product-to-sum identity, this implies:

$\displaystyle \cos(2\theta_1+\theta_3)=\cos(2\theta_2+\theta_3)$

Given that the angles $\displaystyle \theta_n$ must be distinct, I used:

$\displaystyle \cos(2\theta_1+\theta_3)=\cos(2\theta_2+\theta_3-4\pi)$

From which I obtained:

$\displaystyle \theta_1+\theta_2+\theta_3=2\pi$

However, I was told by someone whose judgement is extremely sound that I was "fudging" here.

Would it also be "fudging" to redefine the angles with:

$\displaystyle -\pi\le\theta_n<\pi$

and then use:

$\displaystyle \cos(2\theta_1+\theta_3)=\cos(-2\theta_2-\theta_3)$

to obtain:

$\displaystyle \theta_1+\theta_2+\theta_3=0$

I admit, this seems to merely be the same "fudge" I used before.(Rofl)

Can anyone offer a hint or nudge in the right direction?

 

[Opalg]

Recently, on another forum, the following problem was posted:

Given three distinct complex numbers:

$\displaystyle z_1,z_2,z_3$

where:

$\displaystyle |z_1|=|z_2|=|z_3|\ne0$

and:

$\displaystyle z_1+z_2z_3,z_2+z_1z_3,z_3+z_1z_2$

are all real, then prove:

$\displaystyle z_1z_2z_3=1$

I began with:

$\displaystyle z_1=re^{\theta_1 i}$

$\displaystyle z_2=re^{\theta_2 i}$

$\displaystyle z_3=re^{\theta_3 i}$

where $\displaystyle 0\le\theta_n<2\pi$

For the 3 expression that are real, I equated the imaginary parts to zero, which gives:

$\displaystyle \sin(\theta_1)+r\sin(\theta_2+\theta_3)=0$

$\displaystyle \sin(\theta_2)+r\sin(\theta_1+\theta_3)=0$

$\displaystyle \sin(\theta_3)+r\sin(\theta_1+\theta_2)=0$

I next solved the first two equations for r, and equated, to find:

$\displaystyle \sin(\theta_1)\sin(\theta_1+\theta_3)=\sin(\theta_2)\sin(\theta_2+\theta_3)$

Using a product-to-sum identity, this implies:

$\displaystyle \cos(2\theta_1+\theta_3)=\cos(2\theta_2+\theta_3)$

Up to that point, your argument is completely watertight. To avoid the fudge in what follows, continue like this.

First, given $\phi, \ \psi$ with $\cos\phi=\cos\psi$, it must be true that either $\psi=\phi+2k\pi$ or $\psi=-\phi+2k\pi$ (for some integer $k$). Applying that to the equation $\cos(2\theta_1+\theta_3)=\cos(2\theta_2+\theta_3)$, you see that either $2\theta_1+\theta_3 = 2\theta_2+\theta_3 + 2k\pi$ or $2\theta_1+\theta_3 = -(2\theta_2+\theta_3) + 2k\pi$. But the first of those alternatives implies that $\theta_1 = \theta_2$, which is not allowed. Therefore the second alternative holds, from which you get $\theta_1 + \theta_2 + \theta_3 = k\pi.$

If you then go back to the equation $\sin(\theta_1)+r\sin(\theta_2+\theta_3)=0$, you can deduce that $k$ must be an even number (otherwise both terms on the left side of that equation would have the same sign). Consequently $\sin(\theta_2+\theta_3) = -\sin\theta_1$, from which it follows that $r=1$. From there it is easy to see that $z_1z_2z_3 = 1.$

I would say that your proof is essentially sound and that the fudge is easily removable. I would prefer to see a more geometric proof of this result, but I don't have any idea of how that might go.

 

[MarkFL]

Thank you!

I did actually consider:

$\displaystyle \theta_1+\theta_2+\theta_3=k\pi$

but I never thought to use one of the 3 equations arising from the conditions on the 3 given expressions being real to show that k must be even. (Smile)

When I fix my argument, I will give you credit for the suggested fix!