In $z^{2} + 2 (1 + 2 i) z - (11 + 2 i) = 0$ find $z$ in form of $a + i b$

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Solution

Given,

$z^{2} + 2 (1 + 2 i) z - (11 + 2 i) = 0$

${(x + y i)}^{2} + 2 (1 + 2 i) (x + y i) - (11 + 2 i) = 0$

$(x^{2} - y^{2} + 2 x - 4 y - 11) + i (- 2 + 2 y + 4 x + 2 x y) = 0 + 0 i$

$[\begin{matrix} x^{2} - y^{2} + 2 x - 4 y - 11 = 0 - 2 + 2 y + 4 x + 2 x y = 0 \end{matrix}]$

solving the above 2 equations, we get,

$[\begin{matrix} x^{2} - y^{2} + 2 x - 4 y - 11 = 0 - 2 + 2 y + 4 x + 2 x y = 0 \end{matrix}] : (\begin{matrix} x = 2, y = - 1 x = - 4, y = - 3 \end{matrix})$

$z = 2 - i, z = - 4 - 3 i$

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