Question

If $z_1,z_2,z_3\in C$ satisfy the system of equations given by $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1,z_1+z_2+z_3=1$ and $z_1{z}_2{z}_3=1$ such that $\mathrm{Im}\left(z_1\right)<\mathrm{Im}\left(z_2\right)<\mathrm{Im}\left(z_3\right)$, then find the value of $\left[\left|z_1+z_3{}^2+z_3{}^3\right|\right]$ where [] denotes the greatest integer function.

Answer 1

Answer: 0002

We have $z_1{z}_2+z_2{z}_3+z_3{z}_1=z_1{z}_2{z}_3\left(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right)={\overline{z}}_1+{\overline{z}}_2+{\overline{z}}_3=\overline{(z_1+z_2+z_3})=1$
$\therefore$ The cubic equation with roots $z_1,z_2$ and $z_3$ in $z$ will be
$\left(z-z_1\right)\left(z-z_2\right)\left(z-z_3\right)=z^3-z^2+z-1=0$
$\Rightarrow (z-1)\left(z^2+1\right)=0$
$z=1,\pm i$
$\because \mathrm{Im}\left(z_1\right)<\mathrm{Im}\left(z_2\right)<\mathrm{Im}\left(z_3\right)$
$\therefore z_1=-i,z_2=1,z_3=i$
Now $\left|z_1+z_2^2+z_3^3\right|=\left|-i+1^2+i^3\right|=|1-2i|=\sqrt{5}$
Hence $\left[\left|z_1+z_2{}^2+z_3{}^3\right|\right]=[\sqrt{5}]=2]$