I have 3 complex number $z_1,\> z_2, \>z_3$ from C and i know that: $$z_1 + z_2 + z_3 = 0 , \>\>\>\>\> z_1^2 + z_2^2 + z_3^2 = 0$$ I have to show that $$|z_1|=|z_2|=|z_3|$$
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Try writing them in the froms $z_1=a_1+ib_2$ .... substitute in the equations that you have, and look at the real part alone and the imaginary part alone (they both equal $0$). Also don't forget to show what you tried please. – Fareed Abi Farraj Nov 13 '19 at 14:44
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@FareedAbiFarraj i've tried that and from first equation i've got that the sum of real and imaginary components are 0 and from the second one i have x_1^2 + x_2^2 + x_3^2 - y_1^2 - y_2^2 - y_3^2 + 2 * i * (x_1 * y_1 + x_2 * y_2 + x_3 * y_3) = 0. I don't see how this should help me.. – Nicoara Nov 13 '19 at 15:02
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Show that $z_i$ are the roots of $z^3 - a = 0$. – Maxim Dec 04 '19 at 16:44
2 Answers
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From the given,
$$ z_1z_2 + z_2z_3 +z_3z_1 =\frac12 \left[(z_1 + z_2 + z_3)^2 - (z_1^2 + z_2^2 + z_3^2) \right]= 0 $$
Substitute lhs with $z_3=-(z_1 + z_2)$,
$$z_1z_2 - (z_1 +z_2)^2 = 0 \implies z_1^2 + z_1z_2 + z_2^2 =0$$
which yields $ {z_1} = \frac12(-1\pm\sqrt3 i){z_2}=e^{i \pi \pm i\frac\pi3}{z_2}\implies \left|{z_1}\right| = |{z_2}|$ and likewise $|{z_1}|= |{z_3}|$. Thus,
$$|z_1|=|z_2|=|z_3|$$
Quanto
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@Nicoara - you’re right; should be 2nd and 3rd quadrants. Thanks for spotting it and corrected – Quanto Nov 16 '19 at 15:43
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@Nicoara Note $\frac{z_1}{z_2} = e^{i a}$. Take module $\left|\frac{z_1}{z_2}\right|^2= e^{i a}e^{-i a}=1$ – Quanto Nov 16 '19 at 16:54
