Question

Let $z$ be a complex number satisfying the equation $z^2-(3+i)z+m+2i=0$, where $m\in R$. Suppose the equation has a real root. Then find the non-real root.

• A. $1-i$
• B. $2(1+i)$
• C. $1+i$
• D. $2(1-i)$

Answer 1

The correct option is C $1+i$

Let $\alpha$ be the real root. Then,
${\alpha }^2-(3+i)\alpha +m+2i=0$
or $({\alpha }^2-3\alpha +m)+i(2-\alpha )=0$
or $\alpha =2$ or $4-6+m=0$ or $m=2$
Product of the roots is $2(1+i)$ with one root as 2.
Hence, the nonreal root is $1+i$.
Ans: C