Question

Let $z$ be a complex number satisfying the equation $z^2-(3+i)z+\lambda +2i=0$, where $\lambda \in R$ and $i=\sqrt{-1}$.Suppose the equation has a real root, then find the non-real root.

Answer 1

Let $\alpha$ be the real root.Then,

${\alpha }^2-(3+i)\alpha +\lambda +2i=0$

$\Rightarrow ({\alpha }^2-3\alpha +\lambda )+i(2-\alpha )=0$

On comparing real and imaginary parts, we get

${\alpha }^2-3\alpha +\lambda =0,2-\alpha =0$

$\Rightarrow {\alpha }^2-3\alpha +\lambda =0,\alpha =2$

$\Rightarrow 2^2-3\times 2+\lambda =0$

$\Rightarrow 4-6+\lambda =0$

$\Rightarrow -2+\lambda =0$

$\Rightarrow \lambda =2$

Let the other root be $\beta$ then

$\alpha +\beta =3+i$

$\Rightarrow 2+\beta =3+i$

$\Rightarrow \beta =3+i-2=1+i$

Hence, the non-real root is $1+i$