Question

If the equation, $z^3+(3+i)z^2-3z-(m+i)=0$, where $m\in R$, has at least one real root, then m can have the value equal to

• A. 1
• B. 2
• C. 3
• D. 5

Answer 1

Answer: (A), (D)

5
1

$z^3+(3+i)z^2-3z-(m+i)=0$ ...(1)
Let 'a' be the real root.
$\Rightarrow a=\overline{a}$
$a^3+(3+i)a^2-3a-(m+i)=0$ ...(2) {Since, a is the root of the eq(1)}
Conjugating eq. (2) we get,
${\overline{a}}^3+(3-i){\overline{a}}^2-3\overline{a}-(m-i)=0$
$a^3+(3-i)a^2-3a-(m-i)=0$ ...(3) {$\because a=\overline{a}$}
Subtracting eq. (2) & eq. (2) we get,
$a^2\left(2i\right)=2i$
$\Rightarrow a=\pm 1$ ...(4)
Multiplying eq. (2) by $(3-i)$ & eq. (3) by $(3+i)$ and then subtracting we get,
$(a^3-3a)(3-i-3-i)=(3-i)(m+i)-(3+i)(m-i)$
$\Rightarrow a^3-3a=m-3$
$\Rightarrow m=1,5$ ...{$\because a=\pm 1$}
Ans: A,D