JEE Main 2005

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Animated Solution for Mathematics - Complex Numbers: 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

Q: Wait, why can't the angle be $\pi$? If they are in opposite directions, wouldn't that change the magnitude?

That is a brilliant observation! If the vectors pointed in opposite directions, the magnitude of the sum would be $|z_1| - |z_2|$ (or vice versa), which is definitely not the sum of their magnitudes. The equality $|z_1 + z_2| = |z_1| + |z_2|$ only holds when they are perfectly aligned in the same direction.

Q: What if one of the complex numbers is zero?

The problem explicitly states that $z_1$ and $z_2$ are non-zero. If one were zero, the argument would be undefined, and the geometric interpretation of a triangle would break down. Always check the constraints!

Q: Is there an algebraic way to prove this without geometry?

Absolutely! You can write $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$. Substituting these into the equation and squaring both sides will lead you to the same conclusion that $\cos(\theta_1 - \theta_2) = 1$, which implies $\theta_1 - \theta_2 = 0$.

Select Answer:

Visualized Solution

The Argand Plane

Let's visualize complex numbers $z_{1}$ and $z_{2}$ as vectors in the Argand plane.

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

The magnitude $∣ z ∣$ represents the length of the vector from the origin.

Vector Addition

To find $z_{1} + z_{2}$ , we use the triangle law of vector addition.

The Resultant Vector

The vector from the origin to the tip of the shifted $z_{2}$ is $z_{1} + z_{2}$ .

Triangle Inequality

In any triangle, the length of one side is strictly less than the sum of the other two sides: $∣ z_{1} + z_{2} ∣ < ∣ z_{1} ∣ + ∣ z_{2} ∣$ .

The Given Condition

The problem states an equality: $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ .

Collapsing the Triangle

For the equality to hold, the triangle must collapse.
The vectors must be collinear and point in the same direction.

Collinear Vectors from Origin

This means both $z_{1}$ and $z_{2}$ , when drawn from the origin, lie on the same ray.

Defining the Arguments

The argument of a complex number is the angle it makes with the positive real axis.
Let $ar g (z_{1}) = θ_{1}$ and $ar g (z_{2}) = θ_{2}$ .

Equating the Arguments

Since $z_{1}$ and $z_{2}$ lie on the exact same ray, their angles must be identical: $θ_{1} = θ_{2}$ .

Calculating the Difference

We need to find $ar g (z_{1}) - ar g (z_{2})$ .
Substituting our values gives $θ_{1} - θ_{1}$ .

Final Conclusion

The difference is exactly $0$ .
Whenever the sum of magnitudes equals the magnitude of the sum, the complex numbers are co-directional.

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The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

Imagine you are standing at the origin of the Argand plane, a vast, two-dimensional canvas where complex numbers come to life as vectors. You have two vectors,

z_{1}

and

z_{2}

, stretching out from where you stand.

The problem asks us to explore the condition

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

. This isn't just an equation; it is a geometric story.

The Vector Dance

When we add two complex numbers, we are essentially performing vector addition. If you take vector

z_{1}

and place the tail of

z_{2}

at the head of

z_{1}

, the vector from the origin to the new tip is

z_{1} + z_{2}

In any general case, these three vectors form a triangle. The Triangle Inequality tells us that the length of one side of a triangle is always strictly less than the sum of the other two sides:

∣ z_{1} + z_{2} ∣ < ∣ z_{1} ∣ + ∣ z_{2} ∣

But our problem gives us an equality. This is the moment of truth.

The Moment of Collapse

For the magnitude of the sum to equal the sum of the magnitudes, the triangle must cease to exist. It must collapse.

The only way for the path from the origin to the tip of

z_{1}

and then to the tip of

z_{2}

to be the same length as the direct path is if the vectors are perfectly collinear and pointing in the same direction. They are not just neighbors; they are overlapping rays.

The Final Revelation

Since

z_{1}

and

z_{2}

lie on the same ray starting from the origin, the angle they make with the positive real axis—their arguments—must be identical. If

ar g (z_{1}) = θ_{1}

and

ar g (z_{2}) = θ_{2}

, then

θ_{1} = θ_{2}

When we calculate the difference

ar g (z_{1}) - ar g (z_{2})

, we are simply subtracting a value from itself. The result is $0$ .

This beautiful, elegant cancellation is the heart of the problem. Whenever you see this condition, remember: it is the universe telling you that these two complex numbers are perfectly in sync, marching in the same direction.

Common Traps & FAQs

Wait, why can't the angle be

π

? If they are in opposite directions, wouldn't that change the magnitude?

That is a brilliant observation! If the vectors pointed in opposite directions, the magnitude of the sum would be

∣ z_{1} ∣ - ∣ z_{2} ∣

(or vice versa), which is definitely not the sum of their magnitudes. The equality

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

only holds when they are perfectly aligned in the same direction.

What if one of the complex numbers is zero?

The problem explicitly states that

z_{1}

and

z_{2}

are non-zero. If one were zero, the argument would be undefined, and the geometric interpretation of a triangle would break down. Always check the constraints!

Is there an algebraic way to prove this without geometry?

Absolutely! You can write

z_{1} = r_{1} e^{i θ_{1}}

and

z_{2} = r_{2} e^{i θ_{2}}

. Substituting these into the equation and squaring both sides will lead you to the same conclusion that

cos (θ_{1} - θ_{2}) = 1

, which implies

θ_{1} - θ_{2} = 0

Animated Solution for Mathematics - Complex Numbers: 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

Select Answer:

Visualized Solution

The Argand Plane

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

Vector Addition

The Resultant Vector

Triangle Inequality

The Given Condition

Collapsing the Triangle

Collinear Vectors from Origin

Defining the Arguments

Equating the Arguments

Calculating the Difference

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Vector Dance

The Moment of Collapse

The Final Revelation

Similar Questions

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

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

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 $z_{1} = 10 + 6 i$ 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 $z_{1}, z_{2}$ are two distinct complex number such that $\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2$ , then

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 Resultant Vector

Triangle Inequality

The Given Condition

Collapsing the Triangle

Collinear Vectors from Origin

Defining the Arguments

Equating the Arguments

Calculating the Difference

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Vector Dance

The Moment of Collapse

The Final Revelation

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 nonzero 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​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​,z2​ are two distinct complex number such that ​21​−z1​zˉ2​z1​−2z2​​​=2, then

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

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

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 $∣ z - 7 - 9 i ∣ = 32$ .

If $z_{1}, z_{2}$ are two distinct complex number such that $\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2$ , then