Consider a quadratic equaiton `az^(2) + bz + c=0`, where a,b,c are complex number. If equaiton has two purely imaginary roots, then which of the following is not ture.

Options

A. `abarb` is purely imaginary
B. `b barc` is purely imaginary
C. `cbara` is purely real
D. none of these

Answer

D

Explanation

Let `z_(1)` (purely imginary ) be a root of the given equation Then,
`z_(1) = - barz_(1)`
and `underline(az_(1)^(2) + bz_(1) + c)=0" "(1)`
`rArr az_(1)^(2) + bz_(1) + c = 0`
`rArr bara barz_(1)^(2) + barb barz_(1) + c = 0`
`rArr bar z bar z_(1)^(2) + bar b bar z_(1) + barc = 0`
`rArr bar a bar z_(1)^(2) - bar b barz_(1) + barc = 0" "(as barz_(1) = - z_(1))" "(2)`
Now Eqs. (1) and (2) must have one common root.
`therefore ( cbara-abarc)^(2) = (barbc+ cbarb) (-abarb - barab)`
Let `z_(1)` and `z_(2)` be two purely imaginary roots. Then,
`barz_(1) = -z_(1), barz_(2) = - z_(2)`
Now , `underline(abarz^(2) + bz + c) = 0" "(3)`
or `az^(2) + bz + c=bar0`
or `bara barz_(20 + barb barz + barc =0`
or `bara z^(2) - barbz + barc = 0" "(4)`
Equations (3) and (4) must be identical as their roots are same.
` therefore (a)/(bara) = -(b)/(barb) =(c)/(barc)`
`rArr abarc = barac, + barab = 0` and `b barc +barbc=0`
. Hence, `barac` is purely real and `abarb` and `bbarc` are purely imaginary .
let `z_(1)` (purely real ) be a root of the given equation . Then ,
`z_(1) = barz_(1)` ltbr gt and ``underline(az_(1)^(2) + bz_(1) + c)= bar0" "(5)`
or `az_(1)^(2) + bz_(1) + c=0`
or `baraz_(1)^(2) + bz_(1) + c = bar0`
or `baraz_(1)^(2) + barb z_(1) + c= 0" "(6)`
Now(5) and (6) must have one common root. Hence,
`(cbara - abarc)^(2) = (b barc - cbarb)(abarb-barab)`