Let z be a complex number satisfying z 3 + 2 z 2 + 4 z ¯ - 8 = 0 , where z ¯ denotes the complex…

Let z be a complex number satisfying z3+2z2+4z¯-8=0, where z¯ 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 z2 is equal to 1 12
Q z-z¯2 is equal to 2 4
R z2+z+z¯2 is equal to 3 8
S z+12 is equal to 4 10
    5 7

The correct option is

  1. P→1 Q→3 R→5 S→4
  2. P→2 Q→1 R→3 S→5
  3. P→2 Q→4 R→5 S→1
  4. P→2 Q→3 R→5 S→4

Solution

Given, $|z|^3 + 2z^2 + 4\bar{z} - 8 = 0 \ldots (1)$ Now on taking conjugate both side we get, $|z|^3 + 2\bar{z}^2 + 4z - 8 = 0 \ldots (2)$ Note that $|\bar{z}| = |z|$ Now subtracting both equations we get, $2(z^2 - \bar{z}^2) + 4(\bar{z} - z) = 0$ $\Rightarrow (z - \bar{z})(2z + 4) = 0$ $\Rightarrow z = \bar{z}$ (not possible) or $4x = 4 \Rightarrow x = 1$ as $z + \bar{z} = 2x$ So, $z = 1 + \lambda i$ $\Rightarrow |z| = \sqrt{1 + \lambda^2}$ and $\bar{z} = 1 - \lambda i$ Now putting the value of $z$, $|z|$ and $\bar{z}$ in given equation we get, $(1 + \lambda^2)^{\frac{3}{2}} + 2(1 - \lambda^2 + 2\lambda i) + 4(1 - \lambda i) - 8 = 0$ $\Rightarrow (1 + \lambda^2)^{\frac{3}{2}} + 2(1 - \lambda^2) = 4$ $\Rightarrow (1 + \lambda^2)^{\frac{3}{2}} = 2(1 + \lambda^2)$ $\Rightarrow (1 + \lambda^2)(\sqrt{1 + \lambda^2} - 2) = 0$ $\Rightarrow \lambda^2 = 3$ Now solving, $P = |z|^2 = 1 + \lambda^2 = 1 + 3 = 4$ $Q = |z - \bar{z}|^2 = |1 + \lambda i - (1 - \lambda i)|^2 = |2\lambda i|^2 = 4\lambda^2 = 12$ $R = |z|^2 + |z + \bar{z}|^2 = 4 + |(1 + \lambda i) + (1 - \lambda i)|^2 = 4 + 4 = 8$ $S = |z + 1|^2 = |1 + \lambda i + 1|^2 = 4 + \lambda^2 = 4 + 3 = 7$ $\therefore P \rightarrow 2, Q \rightarrow 1, R \rightarrow 3, S \rightarrow 5$

Asked in: JEE Advanced 2023 (Paper 1)

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