Let S = { z : z = i ( z 2 + R e ( z ) ) } . Then z S z 2 is equal to [2023]

Question

Let $S = {z \in ℂ : \bar{z} = i (z^{2} + R e (\bar{z}))} .$ Then $\sum_{z \in S} | z |^{2}$ is equal to [2023]

Smart Achievers · Accepted Answer

(1)

$\bar{z} = i (z^{2} + Re (\bar{z}))$

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

$\Rightarrow x = - 2 x y and y = y^{2} - x^{2} - x$

$x = 0 \Rightarrow y = y^{2} \Rightarrow y = 0 or 1$

$y = \frac{- 1}{2} \Rightarrow \frac{- 1}{2} = \frac{1}{4} - x^{2} \Rightarrow x = \frac{1}{2} or \frac{- 3}{2}$

$Possible places of z = 0 + i 0, 0 + i, \frac{1}{2} - \frac{1}{2} i, \frac{- 3}{2} - \frac{i}{2}$

$\Rightarrow \sum | z |^{2} = 0 + 1 + (\frac{1}{4} + \frac{1}{4}) + (\frac{9}{4} + \frac{1}{4}) = 1 + 3 = 4$

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