If the least and the largest real values of $\alpha$, for which the equation $z + \alpha|z – 1| + 2i = 0$ ($z \in C$ and $i = \sqrt{-1}$) has a solution, are p and q respectively; then $4 (p^2 + q^2)$ is equal to

### Analyzing the Setup To solve the equation $z + \alpha|z - 1| + 2i = 0$, we begin by expressing the complex number $z$ in its Cartesian form, $z = x + iy$, where $x, y \in \mathbb{R}$. Substituting this into the original equation yields: $$(x + iy) + \alpha|x + iy - 1| + 2i = 0$$ Grouping the real and imaginary components, we obtain: $$(x + \alpha|x - 1 + iy|) + i(y + 2) = 0$$ ### Isolating the Variables From the imaginary part, we immediately identify the anchor value: $$y + 2 = 0 \implies y = -2$$ Substituting $y = -2$ into the real part, we have: $$x + \alpha\sqrt{(x - 1)^2 + (-2)^2} = 0$$ Simplifying the expression inside the square root, we get: $$x + \alpha\sqrt{x^2 - 2x + 5} = 0$$ ### The Master Equation To isolate the parameter $\alpha$, we rearrange the equation: $$\alpha\sqrt{x^2 - 2x + 5} = -x$$ Squaring both sides results in: $$\alpha^2 = \frac{x^2}{x^2 - 2x + 5}$$ We define the function $f(x) = \frac{x^2}{x^2 - 2x + 5}$ to determine the range of $\alpha^2$. ### Optimization via Calculus To find the range of $f(x)$, we calculate its derivative using the quotient rule: $$f'(x) = \frac{2x(x^2 - 2x + 5) - x^2(2x - 2)}{(x^2 - 2x + 5)^2}$$ Simplifying the numerator leads to: $$f'(x) = \frac{-2x^2 + 10x}{(x^2 - 2x + 5)^2}$$ Setting $f'(x) = 0$ provides the critical points at $x = 0$ and $x = 5$. ### Final Calculation Evaluating the function at these critical points, we find $f(0) = 0$ and $f(5) = \frac{25}{25 - 10 + 5} = \frac{25}{20} = \frac{5}{4}$. As $x \to \pm\infty$, $f(x) \to 1$. Thus, the range of $\alpha^2$ is $[0, \frac{5}{4}]$, which implies: $$-\frac{\sqrt{5}}{2} \leq \alpha \leq \frac{\sqrt{5}}{2}$$ Given $p = -\frac{\sqrt{5}}{2}$ and $q = \frac{\sqrt{5}}{2}$, we calculate the final value: $$4(p^2 + q^2) = 4\left(\frac{5}{4} + \frac{5}{4}\right) = 4\left(\frac{10}{4}\right) = \mathbf{10}$$

JEE Main 2021 (February)

LEVELJEE Advanced

Animated Solution for Mathematics - Complex Numbers: If the least and the largest real values of $α$ , for which the equation $z + α ∣ z -1∣ + 2 i = 0$ ( $z \in C$ and $i = - 1$ ) has a solution, are p and q respectively; then $4 (p^{2} + q^{2})$ is equal to

Q: Is it possible for the denominator to be zero?

Great catch! Always check the domain. The denominator is $x^2 - 2x + 5$. The discriminant is $4 - 20 = -16$, which is negative. This means the quadratic is always positive, so the function is defined for all real $x$.

Enter Numerical Value:

Visualized Solution

Substituting $z = x + i y$

Let $z = x + i y$ , where $x, y \in R$ .
Substitute into the equation: $(x + i y) + α ∣ x + i y - 1∣ + 2 i = 0$ .
Group real and imaginary terms.

Separating Imaginary Parts

The imaginary part must be zero: $y + 2 = 0$ .
Solving this gives: $y = - 2$ .

Formulating the Real Part

The real part must be zero: $x + α ∣ x - 1 + i y ∣ = 0$ .
Substitute $y = - 2$ : $x + α (x - 1)^{2} + (- 2)^{2} = 0$ .
Simplify: $x + α x^{2} - 2 x + 5 = 0$ .

Isolating $α^{2}$

Rearrange to isolate $α$ : $α x^{2} - 2 x + 5 = - x$ .
Square both sides: $α^{2} (x^{2} - 2 x + 5) = x^{2}$ .
Express $α^{2}$ as a function of $x$ : $α^{2} = \frac{x ^{2}}{x ^{2} - 2 x + 5}$ .

Analyzing the Function $f (x)$

Let $f (x) = \frac{x ^{2}}{x ^{2} - 2 x + 5}$ .
To find its range, we differentiate: $f^{'} (x) = \frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 2 x + 5})$ .
Apply the quotient rule: $f^{'} (x) = \frac{2 x ( x ^{2} - 2 x + 5 ) - x ^{2} ( 2 x - 2 )}{( x ^{2} - 2 x + 5 ) ^{2}}$ .
Simplify the numerator: $f^{'} (x) = \frac{- 2 x ^{2} + 10 x}{( x ^{2} - 2 x + 5 ) ^{2}}$ .

Finding Critical Points

Set $f^{'} (x) = 0$ to find critical points.
$- 2 x^{2} + 10 x = 0 ⟹ - 2 x (x - 5) = 0$ .
The critical points are $x = 0$ and $x = 5$ .

Evaluating the Function

At $x = 0$ : $f (0) = \frac{0}{5} = 0$ .
At $x = 5$ : $f (5) = \frac{25}{25 - 10 + 5} = \frac{25}{20} = \frac{5}{4}$ .
Check limits as $x \to \pm \infty$ : $lim_{x \to \pm \infty} f (x) = lim_{x \to \pm \infty} \frac{1}{1 - \frac{2}{x} + \frac{5}{x ^{2}}} = 1$ .

Determining the Range of $α$

The range of $f (x)$ is $[0, \frac{5}{4}]$ .
Therefore, $0 \leq α^{2} \leq \frac{5}{4}$ .
Taking the square root: $- \frac{5}{2} \leq α \leq \frac{5}{2}$ .
Least value $p = - \frac{5}{2}$ , largest value $q = \frac{5}{2}$ .

Final Calculation

We need to find $4 (p^{2} + q^{2})$ .
$p^{2} = (- \frac{5}{2})^{2} = \frac{5}{4}$ .
$q^{2} = (\frac{5}{2})^{2} = \frac{5}{4}$ .
$4 (p^{2} + q^{2}) = 4 (\frac{5}{4} + \frac{5}{4}) = 4 (\frac{10}{4}) = 10$ .

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Analyzing the Setup

To solve the equation

z + α ∣ z - 1∣ + 2 i = 0

, we begin by expressing the complex number

z

in its Cartesian form,

z = x + i y

, where

x, y \in R

. Substituting this into the original equation yields:

(x + i y) + α ∣ x + i y - 1∣ + 2 i = 0

Grouping the real and imaginary components, we obtain:

(x + α ∣ x - 1 + i y ∣) + i (y + 2) = 0

Isolating the Variables

From the imaginary part, we immediately identify the anchor value:

y + 2 = 0 ⟹ y = - 2

Substituting

y = - 2

into the real part, we have:

x + α (x - 1)^{2} + (- 2)^{2} = 0

Simplifying the expression inside the square root, we get:

x + α x^{2} - 2 x + 5 = 0

The Master Equation

To isolate the parameter

α

, we rearrange the equation:

α x^{2} - 2 x + 5 = - x

Squaring both sides results in:

α^{2} = \frac{x ^{2}}{x ^{2} - 2 x + 5}

We define the function

f (x) = \frac{x ^{2}}{x ^{2} - 2 x + 5}

to determine the range of

α^{2}

Optimization via Calculus

To find the range of

f (x)

, we calculate its derivative using the quotient rule:

f^{'} (x) = \frac{2 x ( x ^{2} - 2 x + 5 ) - x ^{2} ( 2 x - 2 )}{( x ^{2} - 2 x + 5 ) ^{2}}

Simplifying the numerator leads to:

f^{'} (x) = \frac{- 2 x ^{2} + 10 x}{( x ^{2} - 2 x + 5 ) ^{2}}

Setting

f^{'} (x) = 0

provides the critical points at

x = 0

and

x = 5

Final Calculation

Evaluating the function at these critical points, we find

f (0) = 0

and

f (5) = \frac{25}{25 - 10 + 5} = \frac{25}{20} = \frac{5}{4}

. As

x \to \pm \infty

f (x) \to 1

Thus, the range of

α^{2}

[0, \frac{5}{4}]

, which implies:

- \frac{5}{2} \leq α \leq \frac{5}{2}

Given

p = - \frac{5}{2}

and

q = \frac{5}{2}

, we calculate the final value:

4 (p^{2} + q^{2}) = 4 (\frac{5}{4} + \frac{5}{4}) = 4 (\frac{10}{4}) = 10

Common Traps & FAQs

Wait, why can't we just use the triangle inequality to bound

∣ z ∣

The triangle inequality is a powerful tool, but it gives you bounds on the modulus of the sum, not the individual components. Here, the equation is coupled with

α

, so we need the exact algebraic relationship between

x

and

α

to find the precise range.

Why did we square both sides? Doesn't that introduce fake solutions?

Squaring is a valid operation here because we are analyzing the range of a function. Since

α^{2}

must be non-negative, squaring ensures we capture the magnitude of

α

correctly. We are not solving for a specific

z

, but finding the set of all possible

α

values.

Is it possible for the denominator to be zero?

Great catch! Always check the domain. The denominator is

x^{2} - 2 x + 5

. The discriminant is

4 - 20 = - 16

, which is negative. This means the quadratic is always positive, so the function is defined for all real

x

Animated Solution for Mathematics - Complex Numbers: If the least and the largest real values of $α$ , for which the equation $z + α ∣ z -1∣ + 2 i = 0$ ( $z \in C$ and $i = - 1$ ) has a solution, are p and q respectively; then $4 (p^{2} + q^{2})$ is equal to

Enter Numerical Value:

Visualized Solution

Substituting $z = x + i y$

Separating Imaginary Parts

Formulating the Real Part

Isolating $α^{2}$

Analyzing the Function $f (x)$

Finding Critical Points

Evaluating the Function

Determining the Range of $α$

Final Calculation

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

Isolating the Variables

The Master Equation

Optimization via Calculus

Final Calculation

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Enter Numerical Value:

Visualized Solution

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=x+iy

Separating Imaginary Parts

Formulating the Real Part

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2

Analyzing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Finding Critical Points

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Final Calculation

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Analyzing the Setup

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The Master Equation

Optimization via Calculus

Final Calculation

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