JEE Advanced 2023 — Paper 1 — MATH-IV solution

JEE Advanced2023Paper 1MATH-IV

Let $z$ be a complex number satisfying $∣ z ∣^{3} + 2 z^{2} + 4 \overset{z}{ˉ} - 8 = 0$ , where $\overset{z}{ˉ}$ denotes the complex conjugate of $z$ . Let the imaginary part of $z$ be nonzero.

Match each entry in List-I to the correct entries in List-II.

List-I	List-II
(P) $∣ z ∣^{2}$ is equal to	(1) 12
(Q) $∣ z - \overset{z}{ˉ} ∣^{2}$ is equal to	(2) 4
(R) $∣ z ∣^{2} + ∣ z + \overset{z}{ˉ} ∣^{2}$ is equal to	(3) 8
(S) $∣ z + 1 ∣^{2}$ is equal to	(4) 10
	(5) 7

A
(P) $\to$ (1); (Q) $\to$ (3); (R) $\to$ (5); (S) $\to$ (4)
B
(P) $\to$ (2); (Q) $\to$ (1); (R) $\to$ (3); (S) $\to$ (5)
C
(P) $\to$ (2); (Q) $\to$ (4); (R) $\to$ (5); (S) $\to$ (1)
D
(P) $\to$ (2); (Q) $\to$ (3); (R) $\to$ (5); (S) $\to$ (4)

Solution

The given equation is $∣ z ∣^{3} + 2 z^{2} + 4 \overset{z}{ˉ} - 8 = 0$ . Taking complex conjugates of both sides yields $∣ z ∣^{3} + 2 \overset{z}{ˉ}^{2} + 4 z - 8 = 0$ .

Subtracting one from the other: $2 (z^{2} - \overset{z}{ˉ}^{2}) + 4 (\overset{z}{ˉ} - z) = 0 \Rightarrow 2 (z - \overset{z}{ˉ}) (z + \overset{z}{ˉ} - 2) = 0$ .

Since $Im (z) \neq = 0$ , $z \neq = \overset{z}{ˉ}$ , hence $z + \overset{z}{ˉ} = 2$ , i.e. $Re (z) = 1$ .

Write $z = 1 + i y$ with $y \neq = 0$ . Then $∣ z ∣^{2} = 1 + y^{2}$ and $∣ z ∣^{3} = (1 + y^{2})^{3/2}$ . Also $z^{2} = (1 - y^{2}) + 2 i y$ and $\overset{z}{ˉ} = 1 - i y$ .

Substitute into the original equation and collect real parts: $(1 + y^{2})^{3/2} + 2 (1 - y^{2}) + 4 - 8 = 0 \Rightarrow (1 + y^{2})^{3/2} = 2 (1 + y^{2})$ .

Therefore $(1 + y^{2})^{1/2} = 2$ , giving $1 + y^{2} = 4$ and $y^{2} = 3$ .

So $∣ z ∣^{2} = 4$ $\to$ (2). And $∣ z - \overset{z}{ˉ} ∣^{2} = ∣2 i y ∣^{2} = 4 y^{2} = 12$ $\to$ (1).

$∣ z + \overset{z}{ˉ} ∣^{2} = 4$ , so $∣ z ∣^{2} + ∣ z + \overset{z}{ˉ} ∣^{2} = 4 + 4 = 8$ $\to$ (3). And $∣ z + 1 ∣^{2} = ∣2 + i y ∣^{2} = 4 + y^{2} = 7$ $\to$ (5).

Mapping: P $\to$ 2, Q $\to$ 1, R $\to$ 3, S $\to$ 5. The correct option is (B).

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