Let the complex number z = x + i y be such that 2 z - 3 i 2 z + i is purely imaginary. If x + y 2 = 0 , then y 4 + y 2 - y is equal to [2023]

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Solution

Q.
Let the complex number $z = x + i y$ be such that $\frac{2 z - 3 i}{2 z + i}$ is purely imaginary. If $x + y^{2} = 0$ , then $y^{4} + y^{2} - y$ is equal to [2023]

1 $\frac{3}{2}$

2 $\frac{2}{3}$

3 $\frac{3}{4}$

4 $\frac{4}{3}$

Ans.
(3)

$Given, \frac{2 z - 3 i}{2 z + i} is purely imaginary.$

$∴$ $\frac{2 z - 3 i}{2 z + i} + \frac{2 \bar{z} + 3 i}{2 \bar{z} - i} = 0$

$\Rightarrow (2 z - 3 i) (2 \bar{z} - i) + (2 \bar{z} + 3 i) (2 z + i) = 0$

$\Rightarrow 4 z \bar{z} - 2 i z - 6 i \bar{z} - 3 + 4 z \bar{z} + 2 i \bar{z} + 6 i z - 3 = 0$

$\Rightarrow 8 z \bar{z} + 4 i z - 4 i \bar{z} - 6 = 0$ $[∵ z = x + i y]$

$\Rightarrow 4 (x^{2} + y^{2}) + 2 i (x + i y) - 2 i (x - i y) - 3 = 0$

$\Rightarrow 4 x^{2} + 4 y^{2} - 4 y - 3 = 0$

$\Rightarrow 4 y^{4} + 4 y^{2} - 4 y - 3 = 0$ $[∵ x + y^{2} = 0 (Given)]$

$\Rightarrow y^{4} + y^{2} - y = \frac{3}{4}$

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