Find the real values of $x$ and $y$ for which the following equation is satisfied.
(i) $(2 + 3 i) x^{2} - (3 - 2 i) y = 2 x - 3 y + 5 i$
(ii) $4 x^{2} + 3 x y + (2 x y - 3 x^{2}) i = 4 y^{2} - (\frac{x^{2}}{2}) + (3 x y - 2 y^{2}) i$

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Solution

Given,

(i)

$(2 + 3 i) x^{2} - (3 - 2 i) y = 2 x - 3 y + 5 i$

$(2 x^{2} - 3 y) + (3 x^{2} + 2 y) i = 2 x - 3 y + 5 i$

$[\begin{matrix} 2 x^{2} - 3 y = 2 x - 3 y 3 x^{2} + 2 y = 5 \end{matrix}]$

solving the above 2 equations, we get,

$[\begin{matrix} 2 x^{2} - 3 y = 2 x - 3 y 3 x^{2} + 2 y = 5 \end{matrix}] \Rightarrow (\begin{matrix} x = 0, & y = \frac{5}{2} x = 1, & y = 1 \end{matrix})$

(ii)

$4 x^{2} + 3 x y + (2 x y - 3 x^{2}) i = 4 y^{2} - \frac{x^{2}}{2} + (3 x y - 2 y^{2}) i$

$8 x^{2} + 6 x y + 2 i (2 x y - 3 x^{2}) = 8 y^{2} - x^{2} + 2 i (3 x y - 2 y^{2})$

$[\begin{matrix} 8 x^{2} + 6 x y = 8 y^{2} - x^{2} 4 x y - 6 x^{2} = 6 x y - 4 y^{2} \end{matrix}]$

solving the above 2 quadratic equations, we get,

$[\begin{matrix} 8 x^{2} + 6 x y = 8 y^{2} - x^{2} 4 x y - 6 x^{2} = 6 x y - 4 y^{2} \end{matrix}] \Rightarrow (\begin{matrix} y = 0, & x = 0 \end{matrix})$

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