Question

If $z_1,z_{2}and{z}_3$ are complex numbers such that $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=1,$ then find the value of $\left|z_1+z_2+z_3\right|.$

Answer 1

Any complex $z$ number can be represented as

$z=r\times (\cos \theta +i\sin \theta )=r\times e^{i\theta }$

if $|z|=1$,then $z=(\cos \theta +i\sin \theta )=e^{i\theta }$

$\frac{1}{z}=e^{-i\theta }=(\cos \theta -i\sin \theta )$

$|\cos \theta +i\sin \theta |=|\cos \theta -i\sin \theta |=1$

$|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|=|(\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3)-i(\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3)|=1$

let $C=(\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_2+\cos {\theta }_3),S=(\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3);$

then,$|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|=|C-iS|$

also ,$|z_1+z_2+z_3|=|C+iS|$

we know that $|C+iS|=|C-iS|=\sqrt{C^2+S^2}$

so,$|z_1+z_2+z_3|=1$