Question

let $|z+\overline{z}|+|z-\overline{z}|=2014$. Then $z$ lies on a

• A. Circle
• B. Straight line
• C. Square
• D. Rectangle

Answer 1

Answer: (A)

Circle

$|z+\overline{z}|+|z-\overline{z}|=2014$

${|z+\overline{z}|}^2=[2014-|z-\overline{z}|{]}^2$

${\left|z\right|}^2+{\left|\overline{z}\right|}^2+2z\overline{z}=(2014{)}^2-{|z-\overline{z}|}^2-2(2014\left)\right(z-\overline{z})$

$z\overline{z}+\overline{z}\overline{z}+2z\overline{z}=(2014{)}^2-{\left|z\right|}^2+{\left|\overline{z}\right|}^2+2z\overline{z}-4028z+4028\overline{z}$

$2{\left|\overline{z}\right|}^2+2{\left|z\right|}^2+4028z-4028\overline{z}=(2014{)}^2$

which is similar to the $e{q}^n$ of circle,

by letting $z=x$ & $y=\overline{z}$

$2x^2+2y^2+4028x-4028y=(2014{)}^2$

which is the equation of a circle with radius $2014$.

option (A) is correct.