JEE Advanced 1987

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: If $z_{1}$ and $z_{2}$ are two nonzero complex numbers such that $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ , then $A r g z_{1} - A r g z_{2}$ is equal to

Q: Is the argument difference always zero?

Yes, for the condition $|z_1 + z_2| = |z_1| + |z_2|$, the vectors must be in the same direction, meaning their angles with the positive Real axis are identical. Thus, $Arg(z_1) - Arg(z_2) = 0$ is the correct result.

Select Answer:

Visualized Solution

The Argand Plane

Complex numbers can be visualized as points or vectors in the Argand Plane.
The horizontal axis is the Real axis ( $R e$ ), and the vertical is the Imaginary axis ( $I m$ ).

Vectors $z_{1}$ and $z_{2}$

Let $z_{1}$ and $z_{2}$ be two non-zero complex numbers.
Their magnitudes $∣ z_{1} ∣$ and $∣ z_{2} ∣$ represent the lengths of these vectors.

Vector Addition

To find $z_{1} + z_{2}$ , we use the Triangle Law of Vector Addition.
Shift $z_{2}$ so its tail sits on the head of $z_{1}$ .
The resultant vector from the origin is $z_{1} + z_{2}$ .

The Triangle Inequality

In any triangle, the sum of the lengths of two sides is always strictly greater than the third side.
Therefore, $∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣$ .
This is the fundamental Triangle Inequality for complex numbers.

The Equality Condition

The problem states a special condition: $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ .
When does the "less than" become "equal to"?
Only when the triangle collapses into a straight line!

Collinear Vectors

For the triangle to collapse, $z_{1}$ and $z_{2}$ must be collinear.
Not just collinear, they must point in the exact same direction.
If they pointed in opposite directions, the sum would be $∣∣ z_{1} ∣ - ∣ z_{2} ∣∣$ .

Concept of Argument

The Argument of a complex number, $A r g (z)$ , is the angle it makes with the positive Real axis.
It tells us the direction of the vector in the Argand plane.

Equating the Arguments

Since $z_{1}$ and $z_{2}$ point in the same direction, they must make the same angle with the Real axis.
Therefore, $A r g (z_{1}) = A r g (z_{2}) = θ$ .

Final Calculation

We need to find the value of $A r g (z_{1}) - A r g (z_{2})$ .
Substitute the equal arguments: $θ - θ = 0$ .
The difference between their arguments is $0$ .

The Way Forward

Conclusion: The condition $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ implies the vectors are parallel and in the same direction.
Final Answer: $A r g (z_{1}) - A r g (z_{2}) = 0$ .
Pro Tip: If $∣ z_{1} + z_{2} ∣ = ∣∣ z_{1} ∣ - ∣ z_{2} ∣∣$ , the vectors are in opposite directions!

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The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

The Geometry of Complex Numbers

Imagine you are standing on a vast, two-dimensional stage known as the Argand plane. This is the home of complex numbers, where every complex number

z = x + i y

is represented as a vector reaching out from the origin.

The horizontal axis is the Real axis, and the vertical axis is the Imaginary axis. When we consider

z_{1}

and

z_{2}

, we are analyzing two distinct vectors, each defined by a magnitude (its length) and an argument (the angle it makes with the positive Real axis).

The Triangle Inequality and the Collapse

Consider the sum

z_{1} + z_{2}

. In the world of vectors, we apply the Triangle Law of Vector Addition: we shift

z_{2}

parallel to itself until its tail rests on the head of

z_{1}

, and the resultant vector from the origin to the new head is

z_{1} + z_{2}

In any triangle formed by these vectors, the sum of the lengths of two sides is always greater than or equal to the third side. This is the fundamental Triangle Inequality:

∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣

Our problem presents a specific condition where the equality holds:

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

. Geometrically, the only way the sum of two sides of a triangle can equal the third side is if the triangle collapses into a single, straight line.

The Direction of Vectors

For this collapse to occur, the vectors

z_{1}

and

z_{2}

must be collinear. Furthermore, they must point in the exact same direction.

If they were pointing in opposite directions, the magnitude of the sum would instead be the difference of their lengths, represented as

∣∣ z_{1} ∣ - ∣ z_{2} ∣∣

. Because they are in the same direction, they are perfectly aligned.

The Final Connection

The argument of a complex number,

A r g (z)

, is the angle it makes with the positive Real axis. If

z_{1}

and

z_{2}

point in the exact same direction, they must make the same angle

θ

with the Real axis.

Therefore, we have

A r g (z_{1}) = θ

and

A r g (z_{2}) = θ

. When we calculate the difference, we find:

A r g (z_{1}) - A r g (z_{2}) = θ - θ = 0

The complexity of the initial equation melts away once you visualize the vectors. The final result is $0$ . Keep this geometric intuition in your toolkit; it is a powerful strategy for solving challenging JEE problems.

Common Traps & FAQs

Why can't we just use

z_{1} = x_{1} + i y_{1}

and

z_{2} = x_{2} + i y_{2}

and solve algebraically?

You absolutely can, but it is like using a sledgehammer to crack a nut! Algebraic expansion will lead to a mess of squares and cross-terms. The geometric approach using the Argand plane reveals the 'why' behind the math, which is the true essence of JEE Advanced preparation.

What if the condition was

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

instead?

That is a fantastic question! If the modulus of the sum equals the difference of the moduli, it implies that the vectors

z_{1}

and

z_{2}

are in opposite directions. In that case, the difference of their arguments would be

π

(or

- π

Is the argument difference always zero?

Yes, for the condition

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

, the vectors must be in the same direction, meaning their angles with the positive Real axis are identical. Thus,

A r g (z_{1}) - A r g (z_{2}) = 0

is the correct result.

Animated Solution for Mathematics - Complex Numbers: If $z_{1}$ and $z_{2}$ are two nonzero complex numbers such that $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ , then $A r g z_{1} - A r g z_{2}$ is equal to

Select Answer:

Visualized Solution

The Argand Plane

Vectors $z_{1}$ and $z_{2}$

Vector Addition

The Triangle Inequality

The Equality Condition

Collinear Vectors

Concept of Argument

Equating the Arguments

Final Calculation

The Way Forward

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Complex Numbers

The Triangle Inequality and the Collapse

The Direction of Vectors

The Final Connection

Similar Questions

If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ , then $a r g z_{1} - a r g z_{2}$ is equal to

Let $z_{1}$ and $z_{2}$ be two distinct complex numbers and let $z = (1 - t) z_{1} + t z_{2}$ for some real number $t$ with $0 < t < 1$ .

If $z_{1}, z_{2}$ are complex numbers such that $Re (z_{1}) = ∣ z_{1} - 1∣$ , $Re (z_{2}) = ∣ z_{2} - 1∣$ , and $arg (z_{1} - z_{2}) = \frac{π}{6}$ , then $Im (z_{1} + z_{2})$ is equal to

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $ar g (z_{1} - z_{2}) = \frac{π}{4}$ and $z_{1}, z_{2}$ satisfy the equation $∣ z - 3∣ = Re (z)$ . Then the imaginary part of $z_{1} + z_{2}$ is equal to .

Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq = z_{2}$ and $∣ z_{1} ∣ = ∣ z_{2} ∣$ . If $z_{1}$ has positive real part and $z_{2}$ has negative imaginary part, then $(z_{1} + z_{2}) / (z_{1} - z_{2})$ may be

For $α, β, z \in C$ and $λ > 1$ , if $λ - 1$ is the radius of the circle $∣ z - α ∣^{2} + ∣ z - β ∣^{2} = 2 λ$ , then $∣ α - β ∣$ is equal to

Let $z_{1} = 10 + 6 i$ and $z_{2} = 4 + 6 i$ . If $z$ is any complex number such that the argument of $\frac{z - z _{1}}{z - z _{2}}$ is $\frac{π}{4}$ , then prove that $∣ z - 7 - 9 i ∣ = 32$ .

If $z_{1} = a + ib$ and $z_{2} = c + i d$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = 1$ and $R e (z_{1} \overset{z}{ˉ}_{2}) = 0$ , then the pair of complex numbers $w_{1} = a + i c$ and $w_{2} = b + i d$ satisfies

Let a circle $C$ in complex plane pass through the points $z_{1} = 3 + 4 i$ , $z_{2} = 4 + 3 i$ and $z_{3} = 5 i$ . If $z (\neq = z_{1})$ is a point on $C$ such that the line through $z$ and $z_{1}$ is perpendicular to the line through $z_{2}$ and $z_{3}$ , then $ar g (z)$ is equal to :

Select Answer:

Visualized Solution

The Argand Plane

Vectors .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​

Vector Addition

The Triangle Inequality

The Equality Condition

Collinear Vectors

Concept of Argument

Equating the Arguments

Final Calculation

The Way Forward

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Complex Numbers

The Triangle Inequality and the Collapse

The Direction of Vectors

The Final Connection

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​ and z2​ are two non-zero complex numbers such that ∣z1​+z2​∣=∣z1​∣+∣z2​∣, then argz1​−argz2​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z1​ and z2​ be two distinct complex numbers and let z=(1−t)z1​+tz2​ for some real number t with 0<t<1.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​,z2​ are complex numbers such that Re(z1​)=∣z1​−1∣, Re(z2​)=∣z2​−1∣, and arg(z1​−z2​)=6π​, then Im(z1​+z2​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For α,β,z∈C and λ>1, if λ−1​ is the radius of the circle ∣z−α∣2+∣z−β∣2=2λ, then ∣α−β∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z1​=10+6i and z2​=4+6i. If z is any complex number such that the argument of z−z2​z−z1​​ is 4π​, then prove that ∣z−7−9i∣=32​.

Vectors $z_{1}$ and $z_{2}$

If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ , then $a r g z_{1} - a r g z_{2}$ is equal to

Let $z_{1}$ and $z_{2}$ be two distinct complex numbers and let $z = (1 - t) z_{1} + t z_{2}$ for some real number $t$ with $0 < t < 1$ .

If $z_{1}, z_{2}$ are complex numbers such that $Re (z_{1}) = ∣ z_{1} - 1∣$ , $Re (z_{2}) = ∣ z_{2} - 1∣$ , and $arg (z_{1} - z_{2}) = \frac{π}{6}$ , then $Im (z_{1} + z_{2})$ is equal to

For $α, β, z \in C$ and $λ > 1$ , if $λ - 1$ is the radius of the circle $∣ z - α ∣^{2} + ∣ z - β ∣^{2} = 2 λ$ , then $∣ α - β ∣$ is equal to

Let $z_{1} = 10 + 6 i$ and $z_{2} = 4 + 6 i$ . If $z$ is any complex number such that the argument of $\frac{z - z _{1}}{z - z _{2}}$ is $\frac{π}{4}$ , then prove that $∣ z - 7 - 9 i ∣ = 32$ .