The solution of the equation $z (¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z - 2 i) = 2 (2 + i)$ are

3 + i, 3 - i

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1 + 3 i, 1 - 3 i

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1 + 3 i, 1 - i

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1 - 3 i, 1 + i

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Solution

The correct option is C $1 + 3 i, 1 - i$
$z (¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z - 2 i) = z ¯ ¯ ¯ z + 2 i z = 2 (2 + 1)$ gives
$x^{2} + y^{2} - 2 y = 4$
and $2 x = 2$ , on equating the real and imaginary parts
$∴ y^{2} - 2 y - 3 = 0$ giving $y = 3, - 1$
The solution are $1 + 3 i$ and $1 - i$ .

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