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Question: Let \(\alpha ,\beta \) be real and \(z\) be a complex number. If \({z^2} + \alpha z + \beta = 0\) ha...

Let α,β\alpha ,\beta be real and zz be a complex number. If z2+αz+β=0{z^2} + \alpha z + \beta = 0 has two distinct roots, on the line Re(z)=1{\text{Re}}\left( z \right) = 1 then it is necessary that
A. β∈[0,1) B. β∈[−1,0) C. ∣β∣=1 D. β∈[1,∞)  {\text{A}}{\text{. }}\beta \in [0,1) \\\ {\text{B}}{\text{. }}\beta \in [ - 1,0) \\\ {\text{C}}{\text{. }}\left| \beta \right| = 1 \\\ {\text{D}}{\text{. }}\beta \in [1,\infty ) \\\

Explanation

Solution

Hint- Here, we will be using the concept of complex conjugate pairs and the formulas for sum and product of the roots of any quadratic equation.
Given, quadratic equation is z2+αz+β=0 →(1){z^2} + \alpha z + \beta = 0{\text{ }} \to {\text{(1)}} where α,β\alpha ,\beta be real and zz be a complex number.
It is also given that the two distinct roots of the above quadratic equation lies on the line Re(z)=1{\text{Re}}\left( z \right) = 1 which means that the real part of both the roots is 1.
As we know that the complex roots corresponding to any quadratic equation occurs in conjugate pairs.
Therefore, let us suppose the two distinct roots of the given quadratic equation be z1=1+i(a){z_1} = 1 + i\left( a \right) z2=1−i(a){z_2} = 1 - i\left( a \right).
Also we know that for any general quadratic equation az2+bz+c=0 →(2)a{z^2} + bz + c = 0{\text{ }} \to {\text{(2)}} with two roots as z1{z_1} and z2{z_2},
Sum of roots of the quadratic equation, z1+z2=−ba →(3){z_1} + {z_2} = \dfrac{{ - b}}{a}{\text{ }} \to {\text{(3)}}
Product of roots of the quadratic equation, z1z2=ca →(4){z_1}{z_2} = \dfrac{c}{a}{\text{ }} \to {\text{(4)}}
On comparing equation (1) with equation (2), we get
a=1, b=αa = 1,{\text{ }}b = \alpha and c=βc = \beta
Using equation (3), we get
z1+z2=−ba⇒1+i(a)+1−i(a)=−α1⇒2=−α⇒α=−2{z_1} + {z_2} = \dfrac{{ - b}}{a} \Rightarrow 1 + i\left( a \right) + 1 - i\left( a \right) = \dfrac{{ - \alpha }}{1} \Rightarrow 2 = - \alpha \Rightarrow \alpha = - 2
Using equation (4), we get
z1z2=ca⇒[1+i(a)][1−i(a)]=β1⇒1+i(a)−i(a)−i2(a2)=β⇒1−i2(a2)=β{z_1}{z_2} = \dfrac{c}{a} \Rightarrow \left[ {1 + i\left( a \right)} \right]\left[ {1 - i\left( a \right)} \right] = \dfrac{\beta }{1} \Rightarrow 1 + i\left( a \right) - i\left( a \right) - {i^2}\left( {{a^2}} \right) = \beta \Rightarrow 1 - {i^2}\left( {{a^2}} \right) = \beta
As we know that i2=−1{i^2} = - 1 ⇒1−(−1)(a2)=β⇒β=1+a2 \Rightarrow 1 - \left( { - 1} \right)\left( {{a^2}} \right) = \beta \Rightarrow \beta = 1 + {a^2}
Also we know that a2⩾0{a^2} \geqslant 0 (always) ⇒1+a2⩾1⇒β⩾1⇒β∈[1,∞) \Rightarrow 1 + {a^2} \geqslant 1 \Rightarrow \beta \geqslant 1 \Rightarrow \beta \in [1,\infty )
Hence, option D is correct.

Note- For any general quadratic equation az2+bz+c=0a{z^2} + bz + c = 0 where zz is a complex number, if one of the root is d+i(e)d + i\left( e \right) then the other root will appear as the complex conjugate of the first root i.e., d−i(e)d - i\left( e \right).