Question

Determine the condition so that the equation $z^2+(a+ib)z+(c+id)=0$ has (i) one root real (ii) both roots equal.

Answer 1

(i) Let $z=k$ ($k$ real) be a root, then
$k^2+(a+ib)k+(c+id)=0$
$\therefore k^2+ak+c=0$ and $bk+d=0$
Eliminating $k$ we have the condition as
$\frac{d^2}{b^2}-\frac{ad}{b}+c=0$ or $d^2-adb+b^{2}c=0$
(ii) Let both the roots be equal and each equal to $p+iq$
$\therefore z^2+(a+ib)z+(c+id)=[z-(p+iq){]}^2$
$=z^2-2z(p+iq)+(p+iq{)}^2$
Comparing, $a+ib=-2(p+iq)$
$\therefore a=-2p$, $b=-2q$ ...$\left(1\right)$
and $c+id=(p+iq{)}^2=p^2-q^2+i2pq$
$\therefore c=p^2-q^2$, $d=2pq$
or $c=\frac{a^2}{4}-\frac{b^2}{4}$ and $d=\frac{ab}{y}$ by $\left(1\right)$
or $a^2-b^2=4c$ and $ab=2d$ are the required conditions.