Question

If the least and the largest real values of \[\alpha \], for which the equation \[z + \alpha \left| {z - 1} \right| + 2i = 0\,\,\left( {z \in C\,{\rm{and}}\,i = \sqrt { - 1} } \right)\] has a solution, are \[p\] and \[q\] respectively. Then find the value of \[4\left( {{p^2} + {q^2}} \right)\].

Answer 1

Hint First we will \[z = x + iy\] in the equation \[z + \alpha \left| {z - 1} \right| + 2i = 0\] and compare the real part and imaginary part of the equation. After comparing real and imaginary, we will get two equations. From the second equation, we will find the discriminant and apply the real root condition that is \[D \ge 0\]. From the inequality, we will find the least and largest value of \[\alpha \]. Then put the value of \[p\] and \[q\] in the expression \[4\left( {{p^2} + {q^2}} \right)\].

Formula used
The discriminant of a quadratic equation \[a{x^2} + bx + c = 0\] is \[{b^2} - 4ac\].
For real roots of a quadratic equation \[a{x^2} + bx + c = 0\], the discriminant must be greater than zero.

Complete step by step solution:
Given equation is \[z + \alpha \left| {z - 1} \right| + 2i = 0\,\,\left( {z \in C\,{\rm{and}}\,i = \sqrt { - 1} } \right)\].
Now putting \[z = x + iy\] in equation \[z + \alpha \left| {z - 1} \right| + 2i = 0\]
\[ \Rightarrow x + iy + \alpha \left| {x + iy - 1} \right| + 2i = 0\] …….(1)
Calculate the magnitude of \[x + iy - 1\].
The magnitude of a complex number \[z = x + iy\] is \[\left| z \right| = \sqrt {{x^2} + {y^2}} \].
The magnitude of \[x + iy - 1\] is \[\sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} \].
Now putting \[\left| {x + iy - 1} \right| = \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} \] in equation (1)
\[ \Rightarrow x + iy + \alpha \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} + 2i = 0\]
Combine real part and complex
\[ \Rightarrow \left( {x + \alpha \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} } \right) + \left( {2 + y} \right)i = 0\]
Compare real and imaginary number
\[x + \alpha \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} = 0\] ….(2)
 and \[\left( {2 + y} \right) = 0\] …..(3)
From equation (3) calculate the value of \[y\]
\[y = - 2\]
Rewrite the equation (2)
 \[x = - \alpha \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} \]
Squaring both sides of the equation
\[{x^2} = {\left( { - \alpha \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} } \right)^2}\]
\[ \Rightarrow {x^2} = {\alpha ^2}\left( {{{\left( {x - 1} \right)}^2} + {y^2}} \right)\]
Apply the formula \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
\[ \Rightarrow {x^2} = {\alpha ^2}\left( {{x^2} - 2x + 1 + {y^2}} \right)\]
\[ \Rightarrow 0 = {\alpha ^2}\left( {{x^2} - 2x + 1 + {y^2}} \right) - {x^2}\]
\[ \Rightarrow \left( {{\alpha ^2} - 1} \right){x^2} - 2x{\alpha ^2} + \left( {1 + {y^2}} \right){\alpha ^2} = 0\]
Compare \[\left( {{\alpha ^2} - 1} \right){x^2} - 2x{\alpha ^2} + \left( {1 + {y^2}} \right){\alpha ^2} = 0\] with \[a{x^2} + bx + c = 0\].
\[a = {\alpha ^2} - 1,b = - 2{\alpha ^2},c = \left( {1 + {y^2}} \right){\alpha ^2}\]
Since \[\left( {{\alpha ^2} - 1} \right){x^2} - 2x{\alpha ^2} + \left( {1 + {y^2}} \right){\alpha ^2} = 0\] has real roots, so the discriminant of the equation must be greater than or equal to zero.
\[{b^2} - 4ac \ge 0\]
Putting \[a = {\alpha ^2} - 1,b = - 2{\alpha ^2},c = \left( {1 + {y^2}} \right){\alpha ^2}\]
\[{\left( { - 2{\alpha ^2}} \right)^2} - 4\left( {{\alpha ^2} - 1} \right)\left( {1 + {y^2}} \right){\alpha ^2} \ge 0\]
\[ \Rightarrow {\alpha ^2}\left[ {4{\alpha ^2} - 4\left( {{\alpha ^2} - 1} \right)\left( {1 + {y^2}} \right)} \right] \ge 0\]
\[ \Rightarrow {\alpha ^2}\left[ {4{\alpha ^2} - \left( {4{\alpha ^2} - 4} \right)\left( {1 + {y^2}} \right)} \right] \ge 0\]
\[ \Rightarrow {\alpha ^2}\left[ {4{\alpha ^2} - \left( {4{\alpha ^2} - 4 + 4{\alpha ^2}{y^2} - 4{y^2}} \right)} \right] \ge 0\]
\[ \Rightarrow {\alpha ^2}\left[ {4{\alpha ^2} - 4{\alpha ^2} + 4 - 4{\alpha ^2}{y^2} + 4{y^2}} \right] \ge 0\]
\[ \Rightarrow {\alpha ^2}\left[ {4 - 4{\alpha ^2}{y^2} + 4{y^2}} \right] \ge 0\]
Putting \[y = - 2\] in the above equation
\[ \Rightarrow {\alpha ^2}\left[ {4 - 4{\alpha ^2}{{\left( { - 2} \right)}^2} + 4{{\left( { - 2} \right)}^2}} \right] \ge 0\]
\[ \Rightarrow {\alpha ^2}\left[ {4 - 16{\alpha ^2} + 16} \right] \ge 0\]
\[ \Rightarrow {\alpha ^2}\left[ { - 16{\alpha ^2} + 20} \right] \ge 0\]
A square of real number must be greater than zero.
\[ \Rightarrow {\alpha ^2} \ge 0\] and \[ - 16{\alpha ^2} + 20 \ge 0\]
Now solving the inequality \[ - 16{\alpha ^2} + 20 \ge 0\]
\[ - 16{\alpha ^2} + 20 \ge 0\]
Multiply -1 both sides of the inequality
\[16{\alpha ^2} - 20 \le 0\]
Divide both sides by 16
\[ \Rightarrow {\alpha ^2} - \dfrac{5}{4} \le 0\]
\[ \Rightarrow {\alpha ^2} - {\left( {\dfrac{{\sqrt 5 }}{2}} \right)^2} \le 0\]
\[ \Rightarrow \alpha - \left( {\dfrac{{\sqrt 5 }}{2}} \right) \le 0\] or \[\alpha + \left( {\dfrac{{\sqrt 5 }}{2}} \right) \ge 0\]
\[ \Rightarrow \alpha \le \left( {\dfrac{{\sqrt 5 }}{2}} \right)\] or \[\alpha \ge - \left( {\dfrac{{\sqrt 5 }}{2}} \right)\]
So, the solution of \[\alpha \] is \[\alpha \in \left[ { - \dfrac{{\sqrt 5 }}{2},\dfrac{{\sqrt 5 }}{2}} \right]\].
Then, the least value of \[\alpha \] is \[ - \dfrac{{\sqrt 5 }}{2}\].
The largest value of \[\alpha \] is \[\dfrac{{\sqrt 5 }}{2}\].
So, the value of \[p\] is \[ - \dfrac{{\sqrt 5 }}{2}\] and the value of \[q\] is \[\dfrac{{\sqrt 5 }}{2}\].
Putting the value of \[p\] and \[q\] in \[4\left( {{p^2} + {q^2}} \right)\].
\[4\left( {{p^2} + {q^2}} \right)\]
\[ = 4\left( {{{\left( { - \dfrac{{\sqrt 5 }}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 5 }}{2}} \right)}^2}} \right)\]
\[ = 4\left( {\dfrac{5}{4} + \dfrac{5}{4}} \right)\]
\[ = 4 \cdot \dfrac{{10}}{4}\]
\[ = 10\]
Hence the value of \[4\left( {{p^2} + {q^2}} \right)\] is 10.

Note: Many students do mistake to solve the inequality \[{\alpha ^2} - {\left( {\dfrac{{\sqrt 5 }}{2}} \right)^2} \le 0\]. They solve it like \[\alpha - \left( {\dfrac{{\sqrt 5 }}{2}} \right) \le 0\] or \[\alpha + \left( {\dfrac{{\sqrt 5 }}{2}} \right) \le 0\] which is incorrect. The correct way is \[\alpha - \left( {\dfrac{{\sqrt 5 }}{2}} \right) \le 0\] or \[\alpha + \left( {\dfrac{{\sqrt 5 }}{2}} \right) \ge 0\].