Mathematics - Complex Number Question with Solution | TestHub

Given, the minimum value $v_0$ of $v={\left|z\right|}^2+{\left|z-3\right|}^2+{\left|z-6i\right|}^2$, $z\in \mathrm{ℂ}$ is attained at $z=z_0$,

Let $z=x+iy$

$v=x^2+y^2+{\left(x-3\right)}^2+y^2+x^2+{\left(y-6\right)}^2$

$=\left(3x^2-6x+9\right)+\left(3y^2-12y+36\right)$

$=3\left(x^2+y^2-2x-4y+15\right)$

$=3\left[{\left(x-1\right)}^2+{\left(y-2\right)}^2+10\right]$

$V_{\min }$ at $z=1+2i=z_0$ and $v_0=30$

So, ${\left|2z_0^2-{\overline{z}}_0^3+3\right|}^2+v_0^2$

$={\left|2{\left(1+2i\right)}^2-{\left(1-2i\right)}^3+3\right|}^2+900$

$={\left|-6+8i-\left(1+8i-6i-12\right)+3\right|}^2+900$

$={\left|8+6i\right|}^2+900$ $=1000$