Complex Numbers — JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 Question with Solution

Let z be 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
(P)

∣ z ∣^{2}

is equal to
(Q)

∣ z - \overset{z}{ˉ} ∣^{2}

is equal to
(R)

∣ z ∣^{2} + ∣ z + \overset{z}{ˉ} ∣^{2}

is equal to
(S)

∣ z + 1 ∣^{2}

is equal to

List-II
(1) 12
(2) 4
(3) 8
(4) 10
(5) 7

The correct option is :

Answer: (B)

Solution

Given:

∣ z ∣^{3} + 2 z^{2} + 4 \overset{z}{ˉ} - 8 = 0

... (1)

Taking conjugate:

∣ z ∣^{3} + 2 \overset{z}{ˉ}^{2} + 4 z - 8 = 0

... (2)

Subtracting (2) from (1):

2 (z^{2} - \overset{z}{ˉ}^{2}) + 4 (\overset{z}{ˉ} - z) = 0

z^{2} - \overset{z}{ˉ}^{2} = 2 (z - \overset{z}{ˉ})

(z + \overset{z}{ˉ}) (z - \overset{z}{ˉ}) = 2 (z - \overset{z}{ˉ})

Since imaginary part is nonzero,

z \neq = \overset{z}{ˉ}

, so:

z + \overset{z}{ˉ} = 2

Let

z = x + i y

, then

2 x = 2

, so

x = 1

and

z = 1 + i y

.

Substituting in (1):

(1 + y^{2})^{3/2} + 2 (1 - y^{2} + 2 i y) + 4 (1 - i y) - 8 = 0

(1 + y^{2})^{3/2} = 2 (1 + y^{2})

1 + y^{2} = 2

y^{2} = 3

, so

y = \pm 3

(P)

∣ z ∣^{2} = 1 + 3 = 4

. P → (2)

(Q)

z - \overset{z}{ˉ} = 2 i y = \pm 2 i 3

, so

∣ z - \overset{z}{ˉ} ∣^{2} = 12

. Q → (1)

(R)

∣ z ∣^{2} + ∣ z + \overset{z}{ˉ} ∣^{2} = 4 + 4 = 8

. R → (3)

(S)

z + 1 = 2 + i 3

, so

∣ z + 1 ∣^{2} = 4 + 3 = 7

. S → (5)

Answer: (B)

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option