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. The additive inverse of non-real root, is

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

Answer 1

Answer: (C)

$-1-i$

$z^2-(3+i)z+m+2i=0$
Let $\alpha$ be the real root.
Then, ${\alpha }^2-(3+i)\alpha +m+2i=0$
$\Rightarrow ({\alpha }^2-3\alpha +m)+i(2-\alpha )=0+i\cdot 0$
$\therefore \alpha -2=0$ and ${\alpha }^2-3\alpha +m=0$
$\Rightarrow \alpha =2$ and $m=2$
Now, product of the roots $=2(1+i)$
with one root as $2$
So, non-real root $=1+i$
$\therefore$ Additive inverse is $-1-i$