JEE Main 2022 — Mathematics Algebra

Given, the minimum value ${v}_{0}$v0​ of $v={|z|}^{2}+{|z-3|}^{2}+{|z-6i|}^{2}$v=∣z∣2+∣z−3∣2+∣z−6i∣2, $z\in \mathbb{C}$z∈C is attained at $z={z}_{0}$z=z0​,

Let $z=x+iy$z=x+iy

$$v={x}^{2}+{y}^{2}+{(x-3)}^{2}+{y}^{2}+{x}^{2}+{(y-6)}^{2}$$v=x2+y2+(x−3)2+y2+x2+(y−6)2

$$=(3{x}^{2}-6x+9)+(3{y}^{2}-12y+36)$$=(3x2−6x+9)+(3y2−12y+36)

$$=3({x}^{2}+{y}^{2}-2x-4y+15)$$=3(x2+y2−2x−4y+15)

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

${V}_{\mathrm{min}}$Vmin​ at $z=1+2i={z}_{0}$z=1+2i=z0​ and ${v}_{0}=30$v0​=30

So, ${|2{z}_{0}^{2}-{\bar{z}}_{0}^{3}+3|}^{2}+{v}_{0}^{2}$∣2z02​−zˉ03​+3∣2+v02​

$$={|2{(1+2i)}^{2}-{(1-2i)}^{3}+3|}^{2}+900$$=∣2(1+2i)2−(1−2i)3+3∣2+900

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

$={|8+6i|}^{2}+900$=∣8+6i∣2+900 $=1000$=1000