JEE MAIN - Mathematics (2022 - 27th July Morning Shift - No. 3)
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 :
1000
1024
1105
1196
Explanation
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 $$|2{(1 + 2i)^2} - {(1 - 2i)^3} + 3{|^2} + 900$$
$$ = |2( - 3 + 4i) - (1 - 8{i^3} - 6i(1 - 2i) + 3{|^2} + 900$$
$$ = | - 6 + 8i - (1 + 8i - 6i - 12) + 3{|^2} + 900$$
$$ = |8 + 6i{|^2} + 900$$
$$ = 1000$$
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