Let $\alpha ,\beta $ be real and z be a complex number. If ${{z}^{2}}+\alpha z+\beta =0$has two distinct roots on the line Re(z) = 1, then it is necessary that:
A. $\beta \in \left( 0,1 \right)$
B. $\beta \in \left( -1,0 \right)$
C. $\left| \beta \right|=0$
D. $\beta \in \left( 1,\infty \right)$

Hint: Find the roots of the given quadratic equation using quadratic formula given of, $\alpha ,\beta \left( Roots \right)=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Where $\alpha ,\beta $ are the roots of $a{{x}^{2}}+bx+c=0$. Equate the real part of the roots to 1 as it is given that the real part of z is equal to 1 (roots lie on Re(z) = 1). After this, solve for discriminant less than zero to get the range for $\beta $ as roots are distinct and complex.

Complete step by step answer:
Given, quadratic equation,
$\Rightarrow {{z}^{2}}+\alpha z+\beta =0$
Roots of this quadratic equation can be found using quadratic formula,
The roots are $z=\dfrac{-\alpha \pm \sqrt{{{\alpha }^{2}}-4\beta }}{2}$
Given roots lie on Re (z) = 1
Real part of root $=\dfrac{-\alpha }{2}$
Equating real part of root to 1, we get,
$\begin{align}
  & \dfrac{-\alpha }{2}=1 \\
 & \Rightarrow \alpha =\left( -2 \right) \\
\end{align}$
Now, as we know, Roots are distinct and complex.
$\therefore $ Discriminant of quadratic equation < 0.
Discriminant of $a{{x}^{2}}+bx+c=0$ is given by $D={{b}^{2}}-4ac$.
$\therefore $ Discriminant of ${{z}^{2}}+\alpha z+\beta =0$ is less than zero.
$D\Rightarrow {{\alpha }^{2}}-4\beta <0$
Putting the value of $\alpha =\left( -2 \right)$, we get,
$\begin{align}
  & \Rightarrow {{\left( -2 \right)}^{2}}-4\beta <0 \\
 & \Rightarrow 4-4\beta <0 \\
 & \Rightarrow 4\left( 1-\beta \right)<0 \\
 & \Rightarrow 1-\beta <0 \\
 & \Rightarrow -\beta <-1 \\
 & \Rightarrow \beta >1 \\
 & \Rightarrow \beta \in \left( 1,\infty \right) \\
\end{align}$

So, the correct answer is “Option D”.

Note: This question can be easily solved by method II also.
Method II:
${{z}^{2}}+\alpha z+\beta =0,\ \ \ \ \ \ \ \ \ \ \alpha ,\beta \in R\ \ \ z\in C$
Let,
$\begin{align}
  & z=x+iy \\
 & given\ \operatorname{Re}\left( z \right)=1 \\
 & \therefore x=1 \\
 & \Rightarrow z=1+iy \\
\end{align}$
Since, the complex roots are conjugate with each other.
$\therefore z=1+iy\ and\ 1-iy$ are two roots of ${{z}^{2}}+\alpha z+\beta =0$
$\begin{align}
  & \Rightarrow \text{Product of roots = }\beta \\
 & \Rightarrow \left( 1+iy \right)\left( 1-iy \right)=\beta \\
 & \therefore \beta =1+{{y}^{2}}\ \ \ \ \ \ \ \ \ \left[ \because {{i}^{2}}=\left( -1 \right) \right] \\
 & \therefore \beta >1 \\
 & \Rightarrow \beta \in \left( 1,\infty \right) \\
\end{align}$