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I wonder if there is a geometric proof or a short proof of the following:

let $z_1,z_2,z_3$ be three complex numbers of modulus $r$. prove that the number $$ \frac{r^4+z_1z_2+z_2z_3+z_3z_1}{z_1+z_2+z_3+z_1z_2z_3} $$ is also of modulus $r$.

I wrote everything in trigonometric form and after a page of calculations the result was clear. I wonder if there is a more elegant and shorter proof, maybe using geometry or another approach.

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If you multiply the denominator by $\bar{z_1}\bar{z_2}\bar{z_3}$, a complex number of modulus $r^3$, you get $$\bar{z_1}z_1 \bar{z_2}\bar{z_3} + \bar{z_2}z_2 \bar{z_1}\bar{z_3} + \bar{z_3}z_3 \bar{z_2}\bar{z_1} + z_1\bar{z_1}z_2\bar{z_2}z_3\bar{z_3}$$ $$= r^2\bar{z_2}\bar{z_3} + r^2\bar{z_1}\bar{z_3} + r^2\bar{z_1}\bar{z_2} + r^6$$ Note this is $r^2$ times the complex conjugate of the numerator, and thus has magnitude $r^2$ times that of the numerator. Thus the original denominator has magnitude ${1 \over r}$ times that of the numerator, and therefore the original fraction has magnitude $r$.

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    $\begingroup$ This is great :) A two lines solution :) $\endgroup$ – Beni Bogosel Jan 5 '12 at 19:25

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