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
\[\left( a \right)\beta \in \left( 0,1 \right)\]
\[\left( b \right)\beta \in \left( -1,0 \right)\]
\[\left( c \right)\left| \beta \right|=1\]
\[\left( d \right)\beta \in \left( 1,\infty \right)\]

Hint: We are given that \[{{z}^{2}}+\alpha z+\beta =0\] has distinct roots that lie on Re (z) = 1. First of all, we will find the root using the quadratic formula, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.\] Then as the root lies on Re (z) = 1, so the real part must be 1. So we equate the real part as 1 and get the value of \[\alpha .\] Then as roots are distinct and complex. So, \[{{b}^{2}}-4ac<0.\] So using this, we get the range of \[\beta .\]

Complete step-by-step answer:
We are given a quadratic equation, \[{{z}^{2}}+\alpha z+\beta =0\] where \[\alpha ,\beta \] are real, z is complex. We know that the root of the quadratic equation \[a{{x}^{2}}+bx+c=0\] is given as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.\]
For our equation, \[{{z}^{2}}+\alpha z+\beta =0\] we have, \[a=1,b=\alpha ,c=\beta \] and x as z. So, we get,
\[z=\dfrac{-\alpha \pm \sqrt{{{\alpha }^{2}}-4\beta }}{2\times 1}\]
\[\Rightarrow z=\dfrac{-\alpha \pm \sqrt{{{\alpha }^{2}}-4\beta }}{2}\]
As root lies on the line Re (z) = 1. So the real part of z is 1. So, we get,
\[\Rightarrow \dfrac{-\alpha }{2}=1\]
Simplifying we get,
\[\Rightarrow -\alpha =2\]
\[\Rightarrow \alpha =-2\]
We have, \[\alpha =-2.\]
Now, we also have that the roots are distinct and complex. So, the discriminant \[{{b}^{2}}-4ac<0\] as \[b=\alpha ,c=\beta ,a=1.\] So,
\[{{b}^{2}}-4ac={{\alpha }^{2}}-4\beta <0\]
\[\Rightarrow {{\alpha }^{2}}-4\beta <0\]
As, \[\alpha =-2\] so we get,
\[\Rightarrow {{\left( -2 \right)}^{2}}-4\beta <0\]
\[\Rightarrow 4-4\beta <0\]
\[\Rightarrow 4<4\beta \]
Cancelling 4, we get,
\[\Rightarrow 1<\beta \]
Therefore, \[\beta \] is greater than 1.
\[\Rightarrow \beta \in \left( 1,\infty \right)\]
Hence, the right option is (d).

Note: We have no numeric value in \[{{z}^{2}}+\alpha z+\beta =0.\] So we will not use completing the square method or the middle term split to find the root. The only option we have is using the quadratic formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.\] In \[z=\dfrac{-\alpha \pm \sqrt{{{\alpha }^{2}}-4\beta }}{2},\] the real part is \[\dfrac{-\alpha }{2}\] because as roots are distinct and complex.