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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 =0 has two distinct roots on the line Re(z) = 1, then it is necessary that
(a)β∈(0,1)\left( a \right)\beta \in \left( 0,1 \right)
(b)β∈(−1,0)\left( b \right)\beta \in \left( -1,0 \right)
(c)∣β∣=1\left( c \right)\left| \beta \right|=1
(d)β∈(1,∞)\left( d \right)\beta \in \left( 1,\infty \right)

Explanation

Solution

We are given that z2+αz+β=0{{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=−b±b2−4ac2a.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 α.\alpha . Then as roots are distinct and complex. So, b2−4ac<0.{{b}^{2}}-4ac<0. So using this, we get the range of β.\beta .

Complete step-by-step answer :
We are given a quadratic equation, z2+αz+β=0{{z}^{2}}+\alpha z+\beta =0 where α,β\alpha ,\beta are real, z is complex. We know that the root of the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 is given as x=−b±b2−4ac2a.x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.
For our equation, z2+αz+β=0{{z}^{2}}+\alpha z+\beta =0 we have, a=1,b=α,c=βa=1,b=\alpha ,c=\beta and x as z. So, we get,
z=−α±α2−4β2×1z=\dfrac{-\alpha \pm \sqrt{{{\alpha }^{2}}-4\beta }}{2\times 1}
⇒z=−α±α2−4β2\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,
⇒−α2=1\Rightarrow \dfrac{-\alpha }{2}=1
Simplifying we get,
⇒−α=2\Rightarrow -\alpha =2
⇒α=−2\Rightarrow \alpha =-2
We have, α=−2.\alpha =-2.
Now, we also have that the roots are distinct and complex. So, the discriminant b2−4ac<0{{b}^{2}}-4ac<0 as b=α,c=β,a=1.b=\alpha ,c=\beta ,a=1. So,
b2−4ac=α2−4β<0{{b}^{2}}-4ac={{\alpha }^{2}}-4\beta <0
⇒α2−4β<0\Rightarrow {{\alpha }^{2}}-4\beta <0
As, α=−2\alpha =-2 so we get,
⇒(−2)2−4β<0\Rightarrow {{\left( -2 \right)}^{2}}-4\beta <0
⇒4−4β<0\Rightarrow 4-4\beta <0
⇒4<4β\Rightarrow 4<4\beta
Cancelling 4, we get,
⇒1<β\Rightarrow 1<\beta
Therefore, β\beta is greater than 1.
⇒β∈(1,∞)\Rightarrow \beta \in \left( 1,\infty \right)
Hence, the right option is (d).

Note : We have no numeric value in z2+αz+β=0.{{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=−b±b2−4ac2a.x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}. In z=−α±α2−4β2,z=\dfrac{-\alpha \pm \sqrt{{{\alpha }^{2}}-4\beta }}{2}, the real part is −α2\dfrac{-\alpha }{2} because as roots are distinct and complex.