Question

The value of $|z{|}^2+|z-3{|}^2+|z-i{|}^2$ is minimum when $z$ equals

• A. $2-\frac{2}{3}i$
• B. $45+3i$
• C. $1+\frac{i}{3}$
• D. $1-\frac{i}{3}$

Answer 1

Answer: (C)

Let $z=x+iy$
$\therefore |z{|}^2+|z-3{|}^2+|z-i{|}^2$
$=|x+iy{|}^2+|(x-3)+iy{|}^2+|x+i(y-1){|}^2$
$=x^2+y^2+(x-3{)}^2+y^2+x^2+(y-1{)}^2$
$=x^2+y^2+x^2-6x+9+y^2+x^2+y^2+1-2y$
$=3x^2+3y^2-6x-2y+10$
$=3\left(x^2-2x+1\right)+3\left(y^2-\frac{2}{3}y+\frac{1}{9}\right)+10-3-\frac{1}{3}$
$=3(x-1{)}^2+3{\left(y-\frac{1}{3}\right)}^2+\frac{20}{3}$
It is minimum, when $x-1=0$ and $y-\frac{1}{3}=0$
$\therefore x=1$ and $y=\frac{1}{3}$
$\therefore z=1+\frac{1}{3}i$