JEE Advanced 1983

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Animated Solution for Mathematics - Complex Numbers: The points $z_{1}, z_{2}, z_{3}, z_{4}$ in the complex plane are the vertices of a parallelogram taken in order if and only if

Q: Why can't I just use the vector subtraction method like $\vec{AB} = \vec{DC}$?

You absolutely can! In the complex plane, a vector from $z_1$ to $z_2$ is represented by $z_2 - z_1$. In a parallelogram, the side vectors are equal and parallel, so $z_2 - z_1 = z_3 - z_4$. If you rearrange this, you get $z_1 + z_3 = z_2 + z_4$. It is the same result, just a different perspective. Both methods are beautiful and mathematically equivalent!

Select Answer:

Visualized Solution

Vertices of a Parallelogram

Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the vertices of a parallelogram taken in order.
In the complex plane, these points represent vectors from the origin.

Property of Diagonals

The most important property of a parallelogram is that its diagonals bisect each other.

Drawing the Diagonals

The diagonals connect opposite vertices: $z_{1}$ to $z_{3}$ , and $z_{2}$ to $z_{4}$ .

The Point of Intersection

Let the point of intersection be $M$ .
Since the diagonals bisect each other, $M$ is the midpoint of both diagonals.

Midpoint of First Diagonal

Using the midpoint formula for the segment joining $z_{1}$ and $z_{3}$ :
$M = \frac{z _{1} + z _{3}}{2}$

Midpoint of Second Diagonal

Similarly, for the segment joining $z_{2}$ and $z_{4}$ :
$M = \frac{z _{2} + z _{4}}{2}$

Equating the Expressions

Since both expressions represent the same point $M$ :
$\frac{z _{1} + z _{3}}{2} = \frac{z _{2} + z _{4}}{2}$

Simplifying the Equation

Multiply both sides by $2$ to eliminate the denominators:
$z_{1} + z_{3} = z_{2} + z_{4}$

Final Conclusion

The required condition for $z_{1}, z_{2}, z_{3}, z_{4}$ to form a parallelogram is:
$z_{1} + z_{3} = z_{2} + z_{4}$
This matches the second option.

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The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

In the complex plane, we represent four complex numbers

z_{1}, z_{2}, z_{3},

and

z_{4}

as the vertices of a quadrilateral. For these points to form a parallelogram, they must satisfy specific geometric constraints regarding their orientation and symmetry.

The most fundamental property of a parallelogram is that its diagonals bisect each other. This means the midpoint of the diagonal connecting

z_{1}

and

z_{3}

must coincide with the midpoint of the diagonal connecting

z_{2}

and

z_{4}

The Magic of Diagonals

Let

M

be the point of intersection of the diagonals. Using the midpoint formula in the complex plane, we define the midpoint of the first diagonal as:

M = \frac{z _{1} + z _{3}}{2}

Similarly, the midpoint of the second diagonal is given by:

M = \frac{z _{2} + z _{4}}{2}

The Algebraic Bridge

Since both expressions represent the same point

M

, we equate them to establish the condition for the parallelogram:

\frac{z _{1} + z _{3}}{2} = \frac{z _{2} + z _{4}}{2}

By multiplying both sides of the equation by

2

, we arrive at the elegant master equation:

$z_{1} + z_{3} = z_{2} + z_{4}$

This result is the definitive algebraic condition for four complex numbers to form a parallelogram. It demonstrates that the sum of the complex numbers at opposite vertices must be equal.

Why This Matters for JEE

In the JEE Advanced examination, geometric properties are frequently presented as algebraic challenges. Recognizing that $z_{1} + z_{3} = z_{2} + z_{4}$ is a direct consequence of diagonal bisection allows you to bypass complex coordinate geometry.

This identity transforms daunting geometric visualizations into straightforward algebraic manipulations. Keep this tool in your toolkit to solve problems involving complex loci and transformations with speed and precision.

Remember, every equation tells a story, and every geometric shape possesses a unique mathematical soul. Continue exploring these connections to master the complex plane.

Common Traps & FAQs

Wait, does the order of vertices matter? What if I labeled them

z_{1}, z_{3}, z_{2}, z_{4}

Yes, absolutely! The problem states the vertices are taken in order. If you label them

z_{1}, z_{2}, z_{3}, z_{4}

around the perimeter, the diagonals connect

z_{1}

z_{3}

and

z_{2}

z_{4}

. If you change the order, you change which points are opposite, and the formula

z_{1} + z_{3} = z_{2} + z_{4}

would no longer hold for those specific labels. Always draw the figure first!

Why can't I just use the vector subtraction method like

A B = D C

You absolutely can! In the complex plane, a vector from

z_{1}

z_{2}

is represented by

z_{2} - z_{1}

. In a parallelogram, the side vectors are equal and parallel, so

z_{2} - z_{1} = z_{3} - z_{4}

. If you rearrange this, you get

z_{1} + z_{3} = z_{2} + z_{4}

. It is the same result, just a different perspective. Both methods are beautiful and mathematically equivalent!

Is this property true for all quadrilaterals, or just parallelograms?

Great question! No, it is not true for all quadrilaterals. This property is unique to parallelograms. For a general quadrilateral, the diagonals do not necessarily bisect each other. The bisection property is the defining characteristic that makes a quadrilateral a parallelogram.

Animated Solution for Mathematics - Complex Numbers: The points $z_{1}, z_{2}, z_{3}, z_{4}$ in the complex plane are the vertices of a parallelogram taken in order if and only if

Select Answer:

Visualized Solution

Vertices of a Parallelogram

Property of Diagonals

Drawing the Diagonals

The Point of Intersection

Midpoint of First Diagonal

Midpoint of Second Diagonal

Equating the Expressions

Simplifying the Equation

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Magic of Diagonals

The Algebraic Bridge

Why This Matters for JEE

Similar Questions

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

If the four complex numbers $z, \overset{z}{ˉ}, z - 2 Re (z)$ and $\overset{z}{ˉ} - 2 Re (\overset{z}{ˉ})$ represent the vertices of a square of side 4 units in the Argand plane, then $∣ z ∣$ is equal to :

Prove that the complex numbers $z_{1}, z_{2}$ and the origin form an equilateral triangle only if $z_{1}^{2} + z_{2}^{2} - z_{1} z_{2} = 0$ .

If $a, b, c$ and $u, v, w$ are complex numbers representing the vertices of two triangles such that $c = (1 - r) a + r b$ and $w = (1 - r) u + r v$ , where $r$ is a complex number, then the two triangles

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 $0 < t < 1$ .

Match the statements in Column I with those in Column II. [Note : Here $z$ takes values in the complex plane and $I m z$ and $R ez$ denote , respectively, the imaginary part and the real part of $z$ . ]

List-I

List-II

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

For a non-zero complex number $z$ , let $ar g (z)$ denote the principal argument with $- π < ar g (z) \leq π$ . Then, which of the following statement(s) is (are) FALSE ?

The number of complex numbers $z$ such that $∣ z - 1∣ = ∣ z + 1∣ = ∣ z - i ∣$ equals

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

Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The points z1​,z2​,z3​,z4​ in the complex plane are the vertices of a parallelogram taken in order if and only if

Select Answer:

Visualized Solution

Vertices of a Parallelogram

Property of Diagonals

Drawing the Diagonals

The Point of Intersection

Midpoint of First Diagonal

Midpoint of Second Diagonal

Equating the Expressions

Simplifying the Equation

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Magic of Diagonals

The Algebraic Bridge

Why This Matters for JEE

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the four complex numbers z,zˉ,z−2Re(z) and zˉ−2Re(zˉ) represent the vertices of a square of side 4 units in the Argand plane, then ∣z∣ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that the complex numbers z1​,z2​ and the origin form an equilateral triangle only if z12​+z22​−z1​z2​=0.

.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.

List-I

List-II

.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; } For a non-zero complex number z, let arg(z) denote the principal argument with −π<arg(z)≤π. Then, which of the following statement(s) is (are) FALSE ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of complex numbers z such that ∣z−1∣=∣z+1∣=∣z−i∣ equals

.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

Animated Solution for Mathematics - Complex Numbers: The points $z_{1}, z_{2}, z_{3}, z_{4}$ in the complex plane are the vertices of a parallelogram taken in order if and only if

If the four complex numbers $z, \overset{z}{ˉ}, z - 2 Re (z)$ and $\overset{z}{ˉ} - 2 Re (\overset{z}{ˉ})$ represent the vertices of a square of side 4 units in the Argand plane, then $∣ z ∣$ is equal to :

Prove that the complex numbers $z_{1}, z_{2}$ and the origin form an equilateral triangle only if $z_{1}^{2} + z_{2}^{2} - z_{1} z_{2} = 0$ .

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 $z_{1}$ 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

For a non-zero complex number $z$ , let $ar g (z)$ denote the principal argument with $- π < ar g (z) \leq π$ . Then, which of the following statement(s) is (are) FALSE ?

The number of complex numbers $z$ such that $∣ z - 1∣ = ∣ z + 1∣ = ∣ z - i ∣$ equals

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