Answer
Verified
427.5k+ views
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., $d - i\left( e \right)$.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Two charges are placed at a certain distance apart class 12 physics CBSE
Difference Between Plant Cell and Animal Cell
What organs are located on the left side of your body class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The planet nearest to earth is A Mercury B Venus C class 6 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE