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

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

Explanation

Solution

Find the roots of the given quadratic equation using quadratic formula given of, α,β(Roots)=−b±b2−4ac2a\alpha ,\beta \left( Roots \right)=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Where α,β\alpha ,\beta are the roots of ax2+bx+c=0a{{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 β\beta as roots are distinct and complex.

Complete step by step answer:
Given, quadratic equation,
⇒z2+αz+β=0\Rightarrow {{z}^{2}}+\alpha z+\beta =0
Roots of this quadratic equation can be found using quadratic formula,
The roots are z=−α±α2−4β2z=\dfrac{-\alpha \pm \sqrt{{{\alpha }^{2}}-4\beta }}{2}
Given roots lie on Re (z) = 1
Real part of root =−α2=\dfrac{-\alpha }{2}
Equating real part of root to 1, we get,
−α2=1 ⇒α=(−2) \begin{aligned} & \dfrac{-\alpha }{2}=1 \\\ & \Rightarrow \alpha =\left( -2 \right) \\\ \end{aligned}
Now, as we know, Roots are distinct and complex.
∴\therefore Discriminant of quadratic equation < 0.
Discriminant of ax2+bx+c=0a{{x}^{2}}+bx+c=0 is given by D=b2−4acD={{b}^{2}}-4ac.
∴\therefore Discriminant of z2+αz+β=0{{z}^{2}}+\alpha z+\beta =0 is less than zero.
D⇒α2−4β<0D\Rightarrow {{\alpha }^{2}}-4\beta <0
Putting the value of α=(−2)\alpha =\left( -2 \right), we get,
⇒(−2)2−4β<0 ⇒4−4β<0 ⇒4(1−β)<0 ⇒1−β<0 ⇒−β<−1 ⇒β>1 ⇒β∈(1,∞) \begin{aligned} & \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{aligned}

So, the correct answer is “Option D”.

Note: This question can be easily solved by method II also.
Method II:
z2+αz+β=0,          α,β∈R   z∈C{{z}^{2}}+\alpha z+\beta =0,\ \ \ \ \ \ \ \ \ \ \alpha ,\beta \in R\ \ \ z\in C
Let,
z=x+iy given Re⁡(z)=1 ∴x=1 ⇒z=1+iy \begin{aligned} & z=x+iy \\\ & given\ \operatorname{Re}\left( z \right)=1 \\\ & \therefore x=1 \\\ & \Rightarrow z=1+iy \\\ \end{aligned}
Since, the complex roots are conjugate with each other.
∴z=1+iy and 1−iy\therefore z=1+iy\ and\ 1-iy are two roots of z2+αz+β=0{{z}^{2}}+\alpha z+\beta =0
⇒Product of roots = β ⇒(1+iy)(1−iy)=β ∴β=1+y2         [∵i2=(−1)] ∴β>1 ⇒β∈(1,∞) \begin{aligned} & \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{aligned}