If frac{3x + 2iy}{5i - 2} = frac{15}{8x + 3iy | vedclass

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(D) Given that $\frac{3x + 2iy}{-2 + 5i} = \frac{15}{8x + 3iy}$.Cross-multiplying,we get $(3x + 2iy)(8x + 3iy) = 15(-2 + 5i)$.$24x^2 + 9ixy + 16ixy +

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$x = -1, y = -3$ or $x = 1, y = 3$

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(D) Given that $\frac{3x + 2iy}{-2 + 5i} = \frac{15}{8x + 3iy}$.Cross-multiplying,we get $(3x + 2iy)(8x + 3iy) = 15(-2 + 5i)$.$24x^2 + 9ixy + 16ixy + 6i^2y^2 = -30 + 75i$.Since $i^2 = -1$,we have $24x^2 - 6y^2 + 25ixy = -30 + 75i$.Equating real and imaginary parts:$24x^2 - 6y^2 = -30 \implies 4x^2 - y^2 = -5$ (Equation $1$).$25xy = 75 \implies xy = 3$ (Equation $2$).From Equation $2$,$y = \frac{3}{x}$. Substituting into Equation $1$:$4x^2 - (\frac{3}{x})^2 = -5 \implies 4x^2 - \frac{9}{x^2} = -5$.Let $t = x^2$,then $4t - \frac{9}{t} = -5 \implies 4t^2 + 5t - 9 = 0$.$(4t + 9)(t - 1) = 0$. Since $x$ is real,$x^2 = t = 1$,so $x = \pm 1$.If $x = 1$,$y = 3$. If $x = -1$,$y = -3$.Thus,$(x, y) = (1, 3)$ or $(-1, -3)$.

If $\frac{3x + 2iy}{5i - 2} = \frac{15}{8x + 3iy}$,then

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