Find $a$ if $a {(x + i y)}^{2} + (x + i y) + 1 =$ $(x^{2} - y^{2} + x + 1 - 9) + i y (2 x + 1)$

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Solution

$a {(x + i y)}^{2} + (x + i y) + 1 (Given)$
$\Rightarrow a (x^{2} + i^{2} y^{2} + 2 x y i) + x + i y + 1 = 0$
$\Rightarrow a x^{2} - a y^{2} + 2 a x y i + x + i y + 1 = 0$
$\Rightarrow (a x^{2} - a y^{2} + x + 1) + (2 a x y + y) i = 0 . . . . . . . . . . (1)$
$(x^{2} - y^{2} + x + 1 - 9) + i y (2 x + 1) = 0 . . . . . . . . . . (2)$
From ${e q}^{n} (1) & (2)$ , we have
$(a x^{2} - a y^{2} + x + 1) + (2 a x y + y) i = (x^{2} - y^{2} + x + 1 - 9) + i y (2 x + 1)$
On comparing real and imaginary parts, we have
$a x^{2} - a y^{2} + x + 1 = x^{2} - y^{2} + x + 1 - 9$
we get $a = 1$

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