Let the minimum value v_0 of v=|z|^2+|z-3|^2+|z-6i|^2, z C, be attained at z=z_0.

Question

Let the minimum value $v_0$ of $v=|z|^2+|z-3|^2+|z-6i|^2,\ z\in\mathbb C$, be attained at $z=z_0$. Then $\left|2z_0^2-\bar z_0^3+3\right|^2+v_0^2$ is equal to

Accepted Answer

Let $z=x+iy$:
$$v=x^2+y^2+(x-3)^2+y^2+x^2+(y-6)^2=3x^2-6x+9+3y^2-12y+36=3(x-1)^2+3(y-2)^2+30.$$

Minimum at $z_0=1+2i$, with $v_0=30$.

$$z_0^2=(1+2i)^2=-3+4i, \qquad \bar z_0=1-2i,\quad \bar z_0^3=(1-2i)^3=-11+2i.$$

$$2z_0^2-\bar z_0^3+3=(-6+8i)-(-11+2i)+3=8+6i.$$

$|8+6i|^2=64+36=100.$

$100+v_0^2=100+900=1000.$

Solutions · 1