Let S = { z : 4 z 2 + z = 0 } . Then z S | z | 2 is equal to: [2026]

Question

Let $S = {z \in ℂ : 4 z^{2} + \bar{z} = 0} .$ Then $\sum_{z \in S} | z |^{2}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(4)

$4 z^{2} + \bar{z} = 0$

$Let z = x + i y$

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

$4 x^{2} - 4 y^{2} + 8 x y i + x - i y = 0$

$4 x^{2} - 4 y^{2} + x = 0 & y (8 x - 1) = 0$

$\Rightarrow y = 0 or x = \frac{1}{8}$

$If y = 0, 4 x^{2} + x = 0$

$x = 0, \frac{- 1}{4}$

$∴$ $z_{1} = 0 + 0 i, | z_{1} |^{2} = 0$

$z_{2} = 0 - \frac{1}{4} i, | z_{2} |^{2} = \frac{1}{16}$

$If x = \frac{1}{8},$

$4 \cdot \frac{1}{64} - 4 y^{2} + \frac{1}{8} = 0$

$\Rightarrow 4 y^{2} = \frac{3}{16} \Rightarrow y = \pm \frac{\sqrt{3}}{8}$

$∴$ $z_{3} = \frac{1}{8} + \frac{\sqrt{3}}{8} i, | z_{3} |^{2} = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}$

$z_{4} = \frac{1}{8} - \frac{\sqrt{3}}{8} i, | z_{4} |^{2} = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}$

$∴$ $\sum_{i = 1}^{n} | z_{i} |^{2} = 0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$

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