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