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JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Complex Numbers

Let the minimum value

v_{0}

of

v = ∣ z ∣^{2} + ∣ z - 3 ∣^{2} + ∣ z - 6 i ∣^{2}

,

z \in C

is attained at

z = z_{0}

. Then

2 z_{0}^{2} - \overset{z}{ˉ}_{0}^{3} + 3^{2} + v_{0}^{2}

is equal to

(A)1000✓ Correct
(B)1024
(C)1105
(D)1196

Answer: (A)

Solution

The minimum of

v = ∣ z ∣^{2} + ∣ z - 3 ∣^{2} + ∣ z - 6 i ∣^{2}

occurs at the centroid of the points

0, 3, 6 i

, which is

z_{0} = \frac{0 + 3 + 0}{3} + i \frac{0 + 0 + 6}{3} = 1 + 2 i

.

Compute

v_{0} = ∣1 + 2 i ∣^{2} + ∣1 + 2 i - 3 ∣^{2} + ∣1 + 2 i - 6 i ∣^{2} = 5 + ∣ - 2 + 2 i ∣^{2} + ∣1 - 4 i ∣^{2} = 5 + 8 + 17 = 30

.

Now compute

2 z_{0}^{2} - \overset{z}{ˉ}_{0}^{3} + 3

where

z_{0} = 1 + 2 i

and

\overset{z}{ˉ}_{0} = 1 - 2 i

:

z_{0}^{2} = (1 + 2 i)^{2} = 1 - 4 + 4 i = - 3 + 4 i

\overset{z}{ˉ}_{0}^{3} = (1 - 2 i)^{3} = 1 - 6 i + 12 i^{2} - 8 i^{3} = 1 - 6 i - 12 + 8 i = - 11 + 2 i

2 z_{0}^{2} - \overset{z}{ˉ}_{0}^{3} + 3 = 2 (- 3 + 4 i) - (- 11 + 2 i) + 3 = - 6 + 8 i + 11 - 2 i + 3 = 8 + 6 i

∣8 + 6 i ∣^{2} = 64 + 36 = 100

Therefore

100 + 900 = 1000

. Answer: (A)

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option

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