Why do we reject the root $2 + \sqrt{3}$?

Always keep your eyes on the constraints! The problem explicitly states that $a$ and $b$ must be between 0 and 1. Since $\sqrt{3} \approx 1.732$, the value $2 + \sqrt{3} \approx 3.732$, which is clearly outside the allowed domain. We only keep the root that fits the physical reality of the problem.

JEE Advanced 1989

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: If $a, b, c$ , are the numbers between 0 and 1 such that the points $z_{1} = a + i, z_{2} = 1 + bi$ and $z_{3} = 0$ form an equilateral triangle, then $a = \dots$ and $b = \dots$

Q: Wait, why does $|z_1| = |z_2|$ imply $a=b$?

Great observation! Since $|z_1|^2 = a^2 + 1^2 = a^2 + 1$ and $|z_2|^2 = 1^2 + b^2 = 1 + b^2$, equating them gives $a^2 + 1 = 1 + b^2$, which simplifies to $a^2 = b^2$. Because the problem states $a$ and $b$ are between 0 and 1 (positive), we must have $a = b$.

Q: Why can't we just use rotation to solve this?

You absolutely can! You could use the property that $z_1 = z_2 e^{i\pi/3}$ or $z_2 = z_1 e^{i\pi/3}$. However, for this specific problem, the modulus method is often faster and less prone to complex algebraic errors involving trigonometry. Both paths lead to the same beautiful truth!

Visualized Solution

Visualizing the Points in the Argand Plane

Given points: $z_{1} = a + i$ , $z_{2} = 1 + bi$ , and $z_{3} = 0$ .
Constraint: $0 < a, b < 1$ .
The points form an equilateral triangle in the Argand plane.

Applying the Equilateral Property

For an equilateral triangle, all sides are equal.
$∣ z_{1} - z_{3} ∣ = ∣ z_{2} - z_{3} ∣ = ∣ z_{1} - z_{2} ∣$

Equating Distances from Origin

Using $z_{3} = 0$ , the distances simplify.
$∣ z_{1} - 0∣ = ∣ z_{2} - 0∣ \Rightarrow ∣ z_{1} ∣ = ∣ z_{2} ∣$

Calculating Moduli Squared

$∣ z_{1} ∣^{2} = a^{2} + 1^{2} = a^{2} + 1$
$∣ z_{2} ∣^{2} = 1^{2} + b^{2} = 1 + b^{2}$
Equating them: $a^{2} + 1 = 1 + b^{2}$

Deducing the Relation Between $a$ and $b$

$a^{2} + 1 = 1 + b^{2} \Rightarrow a^{2} = b^{2}$
Since $a, b > 0$ , we take the positive root.
Result: $a = b$

Setting up the Third Side

We must also use the third side of the triangle.
$∣ z_{1} ∣^{2} = ∣ z_{1} - z_{2} ∣^{2}$

Distance Formula for $z_{1}$ and $z_{2}$

$z_{1} - z_{2} = (a - 1) + i (1 - b)$
$∣ z_{1} - z_{2} ∣^{2} = (a - 1)^{2} + (1 - b)^{2}$

Substituting $b = a$

Substitute $b = a$ into the distance squared.
$∣ z_{1} - z_{2} ∣^{2} = (a - 1)^{2} + (1 - a)^{2}$

Simplifying the Distance Expression

Notice that $(1 - a)^{2} = (a - 1)^{2}$ .
$∣ z_{1} - z_{2} ∣^{2} = (a - 1)^{2} + (a - 1)^{2} = 2 (a - 1)^{2}$
Equate to $∣ z_{1} ∣^{2}$ : $a^{2} + 1 = 2 (a - 1)^{2}$

Expanding the Equation

Expand the right side: $2 (a - 1)^{2} = 2 (a^{2} - 2 a + 1)$
$a^{2} + 1 = 2 a^{2} - 4 a + 2$

Forming the Quadratic Equation

Rearrange terms to one side:
$2 a^{2} - a^{2} - 4 a + 2 - 1 = 0$
$a^{2} - 4 a + 1 = 0$

Solving the Quadratic Equation

Use the quadratic formula: $a = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( 1 ) ( 1 )}{2 ( 1 )}$
$a = \frac{4 \pm 16 - 4}{2} = \frac{4 \pm 12}{2}$
$a = \frac{4 \pm 2 3}{2} = 2 \pm 3$

Applying Constraints

We have two possible roots: $a = 2 + 3$ and $a = 2 - 3$ .
Recall the constraint: $0 < a < 1$ .
Since $3 \approx 1.732$ , $2 + 3 \approx 3.732 > 1$ .
Therefore, we must reject $2 + 3$ .

Final Answer

The only valid root is $a = 2 - 3$ .
Since $a = b$ , we also have $b = 2 - 3$ .
Final Answer: $a = 2 - 3, b = 2 - 3$

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The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

Imagine standing in the vast, open space of the Argand plane. We are given three points:

z_{1} = a + i

z_{2} = 1 + bi

, and the origin

z_{3} = 0

These points form the vertices of an equilateral triangle. In the world of complex numbers, this geometric constraint implies the equality of distances between all vertices.

The Power of Modulus

For an equilateral triangle, the distance between any two vertices must be identical. Mathematically, this is expressed as:

∣ z_{1} - z_{3} ∣ = ∣ z_{2} - z_{3} ∣ = ∣ z_{1} - z_{2} ∣

Since

z_{3} = 0

, the first two terms simplify to

∣ z_{1} ∣ = ∣ z_{2} ∣

. We calculate the squared modulus to avoid square roots, using

∣ z ∣^{2} = x^{2} + y^{2}

∣ z_{1} ∣^{2} = a^{2} + 1^{2} = a^{2} + 1

∣ z_{2} ∣^{2} = 1^{2} + b^{2} = 1 + b^{2}

Equating these, we find

a^{2} + 1 = 1 + b^{2}

, which leads to

a^{2} = b^{2}

. Given the constraint

0 < a, b < 1

, we conclude that

a = b

The Algebraic Battle

Now that we know

a = b

, we utilize the third side of the triangle. The distance between

z_{1}

and

z_{2}

must equal the distance from the origin, so we set

∣ z_{1} ∣^{2} = ∣ z_{1} - z_{2} ∣^{2}

First, we find the difference:

z_{1} - z_{2} = (a - 1) + i (1 - b)

Since

a = b

, this simplifies to

(a - 1) + i (1 - a)

. The squared modulus is:

∣ z_{1} - z_{2} ∣^{2} = (a - 1)^{2} + (1 - a)^{2} = 2 (a - 1)^{2}

Equating this to our earlier expression for

∣ z_{1} ∣^{2}

a^{2} + 1 = 2 (a - 1)^{2}

Expanding the right side yields

a^{2} + 1 = 2 (a^{2} - 2 a + 1)

, which simplifies to

a^{2} + 1 = 2 a^{2} - 4 a + 2

. Rearranging the terms, we arrive at the quadratic equation:

a^{2} - 4 a + 1 = 0

Final Calculation

We solve for

a

using the quadratic formula:

a = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( 1 ) ( 1 )}{2 ( 1 )}

a = \frac{4 \pm 12}{2} = 2 \pm 3

We have two candidates:

2 + 3

and

2 - 3

. Recalling the constraint

0 < a < 1

, we note that

2 + 3 \approx 3.732

, which is invalid.

The only survivor is

a = 2 - 3

. Since

a = b

, our final result is:

$a = 2 - 3, b = 2 - 3$

Common Traps & FAQs

Wait, why does

∣ z_{1} ∣ = ∣ z_{2} ∣

imply

a = b

Great observation! Since

∣ z_{1} ∣^{2} = a^{2} + 1^{2} = a^{2} + 1

and

∣ z_{2} ∣^{2} = 1^{2} + b^{2} = 1 + b^{2}

, equating them gives

a^{2} + 1 = 1 + b^{2}

, which simplifies to

a^{2} = b^{2}

. Because the problem states

a

and

b

are between 0 and 1 (positive), we must have

a = b

Why can't we just use rotation to solve this?

You absolutely can! You could use the property that

z_{1} = z_{2} e^{iπ /3}

z_{2} = z_{1} e^{iπ /3}

. However, for this specific problem, the modulus method is often faster and less prone to complex algebraic errors involving trigonometry. Both paths lead to the same beautiful truth!

Why do we reject the root

2 + 3

Always keep your eyes on the constraints! The problem explicitly states that

a

and

b

must be between 0 and 1. Since

3 \approx 1.732

, the value

2 + 3 \approx 3.732

, which is clearly outside the allowed domain. We only keep the root that fits the physical reality of the problem.

Animated Solution for Mathematics - Complex Numbers: If $a, b, c$ , are the numbers between 0 and 1 such that the points $z_{1} = a + i, z_{2} = 1 + bi$ and $z_{3} = 0$ form an equilateral triangle, then $a = \dots$ and $b = \dots$

Visualized Solution

Visualizing the Points in the Argand Plane

Applying the Equilateral Property

Equating Distances from Origin

Calculating Moduli Squared

Deducing the Relation Between $a$ and $b$

Setting up the Third Side

Distance Formula for $z_{1}$ and $z_{2}$

Substituting $b = a$

Simplifying the Distance Expression

Expanding the Equation

Forming the Quadratic Equation

Solving the Quadratic Equation

Applying Constraints

Final Answer

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Power of Modulus

The Algebraic Battle

Final Calculation

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Visualized Solution

Visualizing the Points in the Argand Plane

Applying the Equilateral Property

Equating Distances from Origin

Calculating Moduli Squared

Deducing the Relation Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Setting up the Third Side

Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2​

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=a

Simplifying the Distance Expression

Expanding the Equation

Forming the Quadratic Equation

Solving the Quadratic Equation

Applying Constraints

Final Answer

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Power of Modulus

The Algebraic Battle

Final Calculation

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