If 5 + ix^3y^2 and x^3 + y^2 + 6i are conjuga | vedclass

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(C) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $a = c$ and $b = -d$.Given $z_1 = 5 + i(x^3y^2)$ and $z_2 = (x^3 + y^2) +

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$6$

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(C) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $a = c$ and $b = -d$.Given $z_1 = 5 + i(x^3y^2)$ and $z_2 = (x^3 + y^2) + 6i$.Equating real and imaginary parts: $x^3 + y^2 = 5$ and $x^3y^2 = -6$.Let $u = x^3$ and $v = y^2$. Then $u + v = 5$ and $uv = -6$.The quadratic equation $t^2 - (u+v)t + uv = 0$ becomes $t^2 - 5t - 6 = 0$.Solving for $t$: $(t - 6)(t + 1) = 0$,so $t = 6$ or $t = -1$.Since $y^2 = v$ must be non-negative,we have $y^2 = 6$ and $x^3 = -1$,which implies $x = -1$.Then $	an 	heta = \frac{y}{x} = \frac{\pm \sqrt{6}}{-1} = \mp \sqrt{6}$.Therefore,$	an^2 	heta = (\mp \sqrt{6})^2 = 6$.

If $5 + ix^3y^2$ and $x^3 + y^2 + 6i$ are conjugate complex numbers and $\arg(x + iy) = \theta$,then $\tan^2 \theta$ is equal to

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