Question

If $z_1,z_2,z_3$ are unlmodular complex numbers then the greatest value of ${|z_1-z_2|}^2+{|z_2-z_3|}^2{+|z_3-z_1|}^2$ equal to

• A. 3
• B. 6
• C. 9
• D. $\frac{27}{2}$

Answer 1

The correct option is A 3

Given $z_1,z_2,z_3$ are unlmodular complex numbers.

Then $|z_1|=1,|z_2|=1,|z_3|=1$

Now $|z_1-z_2{|}^2+|z_2-z_3{|}^2+|z_3-z_1{|}^2$

$=(z_1-z_2)\left(\overline{z_1-z_2}\right)+(z_2-z_3)\left(\overline{z_2-z_3}\right)+(z_3-z_1)\left(\overline{z_3-z_1}\right)$

$[\because |z{|}^2=z\overline{z}]$

$(z_1-z_2)(\overline{z_1}-{\overline{z}}_2)+(z_2-z_3)(\overline{z_2}-\overline{z_3})+(z_3-z_1)(\overline{z_3}-\overline{z_1})$

$=z_1\overline{z_1}-z_1\overline{z_2}-z_2\overline{z_1}+z_2\overline{z_2}+z_2\overline{z_2}-z_2\overline{z_3}-z_3\overline{z_2}+z_3\overline{z_3}+z_3\overline{z_3}-z_{3\overline{z_1}}-z_1\overline{z_3}+z_1\overline{z_1}$

$=1+2re(-z_1\overline{z_2})+1+1+2Re(-z_2\overline{z_3})+1+1+2Re(-z_3\overline{z_1})+1[\because Re(z)=\frac{z+\overline{z_z}}{2}]$

$\le 6+2|z_1\overline{z_2}|+2|-z_2\overline{z_3}|+2|-z_3\overline{z_1}$

$=6+2|z_1||\overline{z_2}|+2|z_2||\overline{z_3}|+2|-z_3||\overline{z_1}$

$=6+2|z_1||\overline{1z_2}|+2|z_2||\overline{1z_3}|+2|-z_{3||\overline{z_1}}|$

$=6+2+2+2[\because |z|=1-21\because |z|=|z|]$

=12

$ the greatest value of the given exception is 12.