# 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 ${\text{Re}}\left( z \right) = 1$ then it is necessary that

$

{\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 ) \\

$

Answer

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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 ${z^2} + \alpha z + \beta = 0{\text{ }} \to {\text{(1)}}$ where $\alpha ,\beta $ be real and $z$ be a complex number.

It is also given that the two distinct roots of the above quadratic equation lies on the line ${\text{Re}}\left( z \right) = 1$ 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 \[{z_1} = 1 + i\left( a \right)\] \[{z_2} = 1 - i\left( a \right)\].

Also we know that for any general quadratic equation $a{z^2} + bz + c = 0{\text{ }} \to {\text{(2)}}$ with two roots as ${z_1}$ and ${z_2}$,

Sum of roots of the quadratic equation, ${z_1} + {z_2} = \dfrac{{ - b}}{a}{\text{ }} \to {\text{(3)}}$

Product of roots of the quadratic equation, ${z_1}{z_2} = \dfrac{c}{a}{\text{ }} \to {\text{(4)}}$

On comparing equation (1) with equation (2), we get

$a = 1,{\text{ }}b = \alpha $ and $c = \beta $

Using equation (3), we get

${z_1} + {z_2} = \dfrac{{ - b}}{a} \Rightarrow 1 + i\left( a \right) + 1 - i\left( a \right) = \dfrac{{ - \alpha }}{1} \Rightarrow 2 = - \alpha \Rightarrow \alpha = - 2$

Using equation (4), we get

\[{z_1}{z_2} = \dfrac{c}{a} \Rightarrow \left[ {1 + i\left( a \right)} \right]\left[ {1 - i\left( a \right)} \right] = \dfrac{\beta }{1} \Rightarrow 1 + i\left( a \right) - i\left( a \right) - {i^2}\left( {{a^2}} \right) = \beta \Rightarrow 1 - {i^2}\left( {{a^2}} \right) = \beta \]

As we know that ${i^2} = - 1$ \[ \Rightarrow 1 - \left( { - 1} \right)\left( {{a^2}} \right) = \beta \Rightarrow \beta = 1 + {a^2}\]

Also we know that ${a^2} \geqslant 0$ (always) $ \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 $a{z^2} + bz + c = 0$ where $z$ is a complex number, if one of the root is $d + i\left( e \right)$ then the other root will appear as the complex conjugate of the first root i.e., $d - i\left( e \right)$.

Given, quadratic equation is ${z^2} + \alpha z + \beta = 0{\text{ }} \to {\text{(1)}}$ where $\alpha ,\beta $ be real and $z$ be a complex number.

It is also given that the two distinct roots of the above quadratic equation lies on the line ${\text{Re}}\left( z \right) = 1$ 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 \[{z_1} = 1 + i\left( a \right)\] \[{z_2} = 1 - i\left( a \right)\].

Also we know that for any general quadratic equation $a{z^2} + bz + c = 0{\text{ }} \to {\text{(2)}}$ with two roots as ${z_1}$ and ${z_2}$,

Sum of roots of the quadratic equation, ${z_1} + {z_2} = \dfrac{{ - b}}{a}{\text{ }} \to {\text{(3)}}$

Product of roots of the quadratic equation, ${z_1}{z_2} = \dfrac{c}{a}{\text{ }} \to {\text{(4)}}$

On comparing equation (1) with equation (2), we get

$a = 1,{\text{ }}b = \alpha $ and $c = \beta $

Using equation (3), we get

${z_1} + {z_2} = \dfrac{{ - b}}{a} \Rightarrow 1 + i\left( a \right) + 1 - i\left( a \right) = \dfrac{{ - \alpha }}{1} \Rightarrow 2 = - \alpha \Rightarrow \alpha = - 2$

Using equation (4), we get

\[{z_1}{z_2} = \dfrac{c}{a} \Rightarrow \left[ {1 + i\left( a \right)} \right]\left[ {1 - i\left( a \right)} \right] = \dfrac{\beta }{1} \Rightarrow 1 + i\left( a \right) - i\left( a \right) - {i^2}\left( {{a^2}} \right) = \beta \Rightarrow 1 - {i^2}\left( {{a^2}} \right) = \beta \]

As we know that ${i^2} = - 1$ \[ \Rightarrow 1 - \left( { - 1} \right)\left( {{a^2}} \right) = \beta \Rightarrow \beta = 1 + {a^2}\]

Also we know that ${a^2} \geqslant 0$ (always) $ \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 $a{z^2} + bz + c = 0$ where $z$ is a complex number, if one of the root is $d + i\left( e \right)$ then the other root will appear as the complex conjugate of the first root i.e., $d - i\left( e \right)$.

Last updated date: 30th May 2023

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