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

Question

Let the minimum value $v_{0}$ of $$v=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \in \mathbb{C}$$ is attained at  ${ }{z}=z_{0}$. Then $$\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}$$ is equal to :

Accepted Answer

Let $z = x + iy$

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

$$ = (3{x^2} - 6x + 9) + (3{y^2} - 12y + 36)$$

$= 3({x^2} + {y^2} - 2x - 4y + 15)$

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

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

so $$|2{(1 + 2i)^2} - {(1 - 2i)^3} + 3{|^2} + 900$$

$$ = |2( - 3 + 4i) - (1 - 8{i^3} - 6i(1 - 2i) + 3{|^2} + 900$$

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

$= |8 + 6i{|^2} + 900$

$= 1000$

Solutions · 1