Given that the equation z^2 + (p + iq)z + r + | vedclass

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(D) Given the equation $z^2 + (p + iq)z + r + is = 0$ ... $(i)$Let $z = \alpha$ (where $\alpha$ is real) be a root of $(i)$.Then,$\alpha^2 + (p + iq)\

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$pqs = s^2 + q^2r$

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(D) Given the equation $z^2 + (p + iq)z + r + is = 0$ ... $(i)$Let $z = \alpha$ (where $\alpha$ is real) be a root of $(i)$.Then,$\alpha^2 + (p + iq)\alpha + r + is = 0$.This can be written as $(\alpha^2 + p\alpha + r) + i(q\alpha + s) = 0$.Equating the real and imaginary parts to zero,we get:$1) \alpha^2 + p\alpha + r = 0$$2) q\alpha + s = 0$From $(2)$,we have $\alpha = -\frac{s}{q}$.Substituting this into $(1)$,we get $\left(-\frac{s}{q}\right)^2 + p\left(-\frac{s}{q}\right) + r = 0$.$\frac{s^2}{q^2} - \frac{ps}{q} + r = 0$.Multiplying by $q^2$,we get $s^2 - pqs + q^2r = 0$.Therefore,$pqs = s^2 + q^2r$.

Given that the equation $z^2 + (p + iq)z + r + is = 0$,where $p, q, r, s$ are real and non-zero,has a real root,then:

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