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4-1.Complex numbers
hard
જો $p,q,r,s$ એ વાસ્તવિક અને શૂન્યતર હોય તો સમીકરણ ${z^2} + (p + iq)z + r + i\,s = 0 $ ના બીજ વાસ્તવિક હોય તો. . .
A
$pqr = {r^2} + {p^2}s$
B
$prs = {q^2} + {r^2}p$
C
$qrs = {p^2} + {s^2}q$
D
$pqs = {s^2} + {q^2}r$
Solution
(d)Given that ${z^2} + (p + iq)z + r + is = 0$……$(i)$
Let $z = \alpha $ (where $\alpha $ is real) be a root of $(i)$, then
${\alpha ^2} + (p + iq)\alpha + r + is =0$
or ${\alpha ^2} + p\alpha + r + i(q\alpha + s)$$=0$
Equating real and imaginary parts, we have ${\alpha ^2} + p\alpha + r = 0$ and $q\alpha + s = 0$
Eliminating $\alpha ,$we get ${\left( {\frac{{ – s}}{q}} \right)^2} + p\left( {\frac{{ – s}}{q}} \right) + r = 0$
or ${s^2} – pqs + {q^2}r = 0$ or $pqs = {s^2} + {q^2}r$
Standard 11
Mathematics
