Question

The value of ${\left|z\right|}^2+{|z-3|}^2+{|z-i|}^2$ 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)

$1+\frac{i}{3}$

Let $z=x+iy$

$|z{|}^2+|z-3{|}^2+|z-i{|}^2$

$⟹x^2+y^2+(x-3{)}^2+y^2+x^2+(y-1{)}^2$

Differentiating with respect to $x$

$\frac{dy}{dx}=2x+2(x-3)+2x+2y\frac{dy}{dx}+2y\frac{dy}{dx}+2(y-1)\frac{dy}{dx}$

For max value $\frac{dy}{dx}=0$

$2x+2x+2x-6=0$

$⟹x=1$

When $x=1$

$y=1+y^2-2^2+y^2+1+(y-1{)}^2$

$⟹y=6+2y2+(y-1)$

$\frac{dy}{dx}=4y+2(y-1)$

$\frac{dy}{dx}=0$

$6y-2=0$

$⟹y=1/3$

$⟹z=1+\frac{i}{3}$