For a complex number z, the equation z^2+p+iqz+r+is=0 has a real root (where p,q,r,s are non-zero real numbers and i^2=-1), then

Question

MARKS Content Team · Accepted Answer

Correct answer: (D) pqs=s^2+q^2r. Since, the equation has a real root ∴ z=x+iy=x ⇒x^2+p+iqx+r+is=0 ∴x^2+px+r=0 and qx+s=0 Putting x=-s/q in the first equation, we get, pqs=s^2+q^2r which is the required condition.

For a complex number $z$ , the equation $z^{2} + (p + i q) z + r + i s = 0$ has a real root (where $p, q, r, s$ are non-zero real numbers and $i^{2} = - 1$ ), then

Solution

More Complex Number Questions

For a complex number z, the equation z2+p+iqz+r+is=0 has a real root (where p,q,r,s are non-zero real numbers and i2=-1), then

Solution

More Complex Number Questions

For a complex number $z$ , the equation $z^{2} + (p + i q) z + r + i s = 0$ has a real root (where $p, q, r, s$ are non-zero real numbers and $i^{2} = - 1$ ), then