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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.

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