If $z_1,z_2,z_3 \in \Bbb C$ such that $|z_1|=|z_2|= |z_3|=\left|\dfrac{1}{z_1}+\dfrac{1}{z_2}+\dfrac{1}{z_3}\right|=1$, then$|z_1+z_2+z_3|$ is?

Question

If $z_1, z_2, z_3$ are complex numbers such that $|z_1|=|z_2|= |z_3| $= $\left|\dfrac{1}{z_1}+\dfrac{1}{z_2}+\dfrac{1}{z_3}\right| = 1$, then$|z_1 + z_2 + z_3|$ is :

(A) equal to 1(B) less than 1 (C) greater than 3 (D) equal to 3

I am not able to proceed because everything = 1 including the answer which is equal to 1

The whole sum of reciprocal is also in modulas – Samar Imam Zaidi Sep 04 '17 at 02:11 — Samar Imam Zaidi, Sep 04 '17 at 02:11
Done the modification – Samar Imam Zaidi Sep 04 '17 at 02:15 — Samar Imam Zaidi, Sep 04 '17 at 02:15

John Wayland Bales · Accepted Answer · 2017-09-04T03:06:30.413

\begin{eqnarray}1&=&\left|\dfrac{1}{z_1}+\dfrac{1}{z_2}+\dfrac{1}{z_3}\right|\\ &=&\left|\dfrac{1}{z_1}\cdot\frac{\bar z_1}{\bar z_1}+\dfrac{1}{z_2}\cdot\frac{\bar z_2}{\bar z_2}+\dfrac{1}{z_3}\cdot\frac{\bar z_3}{\bar z_3}\right|\\ &=&\left|\dfrac{\bar z_1}{|z_1|^2}+\dfrac{\bar z_2}{|z_2|^2}+\dfrac{\bar z_3}{|\bar z_3|^2}\right|\\ &=&\left|\bar z_1+\bar z_2+\bar z_3\right|\\ &=&\left|\overline{z_1+z_2+z_3}\right|\\ &=&\left|z_1+z_2+z_3\right| \end{eqnarray}

If $z_1,z_2,z_3 \in \Bbb C$ such that $|z_1|=|z_2|= |z_3|=\left|\dfrac{1}{z_1}+\dfrac{1}{z_2}+\dfrac{1}{z_3}\right|=1$, then$|z_1+z_2+z_3|$ is?

1 Answers1