JEE Advanced 1988

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Animated Solution for Mathematics - Complex Numbers: For any two complex numbers $z_{1}, z_{2}$ and any real number $a$ and $b$ , $∣ a z_{1} - b z_{2} ∣^{2} + ∣ b z_{1} + a z_{2} ∣^{2} = \dots$

Q: Why does the conjugate of $a$ remain $a$ in the expansion?

Great observation! Remember that $a$ and $b$ are given as real numbers. For any real number $x$, the conjugate $\bar{x}$ is simply $x$ because the imaginary part is zero. This property is what allows us to treat $a$ and $b$ as constants during the conjugation process, making the algebra much cleaner.

Visualized Solution

Introduction to the Expression

Given expression: $∣ a z_{1} - b z_{2} ∣^{2} + ∣ b z_{1} + a z_{2} ∣^{2}$
Where $z_{1}, z_{2} \in C$ and $a, b \in R$ .

The Modulus-Conjugate Tool

Use the property: $∣ z ∣^{2} = z \cdot \overset{z}{ˉ}$
This allows us to convert modulus terms into algebraic products.

Applying the Property to Term 1

First term: $∣ a z_{1} - b z_{2} ∣^{2} = (a z_{1} - b z_{2}) (\overline{a z_{1} - b z_{2}})$
Since $a, b \in R$ , $\overset{a}{ˉ} = a$ and $\overset{ˉ}{b} = b$ .
$(a z_{1} - b z_{2}) (a \overset{z}{ˉ}_{1} - b \overset{z}{ˉ}_{2})$

Applying the Property to Term 2

Second term: $∣ b z_{1} + a z_{2} ∣^{2} = (b z_{1} + a z_{2}) (\overline{b z_{1} + a z_{2}})$
Applying real number conjugates:
$(b z_{1} + a z_{2}) (b \overset{z}{ˉ}_{1} + a \overset{z}{ˉ}_{2})$

Expanding the First Product

Expanding Term 1: $(a z_{1} - b z_{2}) (a \overset{z}{ˉ}_{1} - b \overset{z}{ˉ}_{2})$
$= a^{2} z_{1} \overset{z}{ˉ}_{1} - ab z_{1} \overset{z}{ˉ}_{2} - ab \overset{z}{ˉ}_{1} z_{2} + b^{2} z_{2} \overset{z}{ˉ}_{2}$
$= a^{2} ∣ z_{1} ∣^{2} + b^{2} ∣ z_{2} ∣^{2} - ab (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2})$

Expanding the Second Product

Expanding Term 2: $(b z_{1} + a z_{2}) (b \overset{z}{ˉ}_{1} + a \overset{z}{ˉ}_{2})$
$= b^{2} z_{1} \overset{z}{ˉ}_{1} + ab z_{1} \overset{z}{ˉ}_{2} + ab \overset{z}{ˉ}_{1} z_{2} + a^{2} z_{2} \overset{z}{ˉ}_{2}$
$= b^{2} ∣ z_{1} ∣^{2} + a^{2} ∣ z_{2} ∣^{2} + ab (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2})$

Combining and Cancelling

Summing the two results:
$(a^{2} ∣ z_{1} ∣^{2} + b^{2} ∣ z_{2} ∣^{2} - ab (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2})) + (b^{2} ∣ z_{1} ∣^{2} + a^{2} ∣ z_{2} ∣^{2} + ab (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2}))$
The cross terms cancel out: $- ab (\dots) + ab (\dots) = 0$

Grouping and Factorizing

Remaining terms: $a^{2} ∣ z_{1} ∣^{2} + b^{2} ∣ z_{1} ∣^{2} + a^{2} ∣ z_{2} ∣^{2} + b^{2} ∣ z_{2} ∣^{2}$
Factor out $∣ z_{1} ∣^{2}$ and $∣ z_{2} ∣^{2}$ :
$(a^{2} + b^{2}) ∣ z_{1} ∣^{2} + (a^{2} + b^{2}) ∣ z_{2} ∣^{2}$
Final factorization: $(a^{2} + b^{2}) (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2})$

Final Takeaway

Key Takeaway: The cross terms cancel out due to the sign difference.
This identity is a generalization of the Parallelogram Law (when $a = 1, b = 1$ ).
Final Result: $(a^{2} + b^{2}) (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2})$

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The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

We are tasked with simplifying the expression

∣ a z_{1} - b z_{2} ∣^{2} + ∣ b z_{1} + a z_{2} ∣^{2}

, where

z_{1}

and

z_{2}

are complex numbers and

a

and

b

are real constants.

At first glance, this appears to be a complex algebraic exercise. However, by applying the fundamental properties of complex numbers, we can reveal a hidden, elegant structure.

The Master Key

The essential tool for this problem is the identity relating the square of a modulus to the product of the complex number and its conjugate:

∣ z ∣^{2} = z \cdot \overset{z}{ˉ}

This identity allows us to translate geometric magnitudes into algebraic products, making the underlying symmetry of the expression visible.

The Algebraic Dance

Let us apply this identity to the first term,

∣ a z_{1} - b z_{2} ∣^{2}

(a z_{1} - b z_{2}) (\overline{a z_{1} - b z_{2}}) = (a z_{1} - b z_{2}) (a \overset{z}{ˉ}_{1} - b \overset{z}{ˉ}_{2})

Since

a

and

b

are real, their conjugates are simply themselves. Expanding the product, we obtain:

a^{2} z_{1} \overset{z}{ˉ}_{1} - ab z_{1} \overset{z}{ˉ}_{2} - ab \overset{z}{ˉ}_{1} z_{2} + b^{2} z_{2} \overset{z}{ˉ}_{2} = a^{2} ∣ z_{1} ∣^{2} + b^{2} ∣ z_{2} ∣^{2} - ab (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2})

Now, we perform the same expansion for the second term,

∣ b z_{1} + a z_{2} ∣^{2}

(b z_{1} + a z_{2}) (\overline{b z_{1} + a z_{2}}) = (b z_{1} + a z_{2}) (b \overset{z}{ˉ}_{1} + a \overset{z}{ˉ}_{2})

Expanding this yields:

b^{2} z_{1} \overset{z}{ˉ}_{1} + ab z_{1} \overset{z}{ˉ}_{2} + ab \overset{z}{ˉ}_{1} z_{2} + a^{2} z_{2} \overset{z}{ˉ}_{2} = b^{2} ∣ z_{1} ∣^{2} + a^{2} ∣ z_{2} ∣^{2} + ab (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2})

The Moment of Cancellation

When we sum the two expanded results, the cross terms

- ab (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2})

and

+ ab (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2})

are identical in magnitude but opposite in sign.

Consequently, these terms vanish completely:

(a^{2} ∣ z_{1} ∣^{2} + b^{2} ∣ z_{2} ∣^{2} - ab (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2})) + (b^{2} ∣ z_{1} ∣^{2} + a^{2} ∣ z_{2} ∣^{2} + ab (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2}))

This leaves us with the sum of the squared terms:

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

Final Synthesis

Finally, we group the terms by factoring out

∣ z_{1} ∣^{2}

and

∣ z_{2} ∣^{2}

(a^{2} + b^{2}) ∣ z_{1} ∣^{2} + (a^{2} + b^{2}) ∣ z_{2} ∣^{2}

Factoring out the common term

(a^{2} + b^{2})

, we arrive at the final, elegant result:

(a^{2} + b^{2}) (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2})

This result is a beautiful generalization of the Parallelogram Law. It serves as a powerful identity to keep in your toolkit for JEE mathematics.

Common Traps & FAQs

Why can't we just treat

z_{1}

and

z_{2}

as simple variables like

x

and

y

and square them directly?

That is a classic trap! Complex numbers are not ordered fields, and the square of a complex number is not the same as the square of its modulus. The modulus squared,

∣ z ∣^{2}

, is specifically defined as

z \cdot \overset{z}{ˉ}

. If you just square the expression

(a z_{1} - b z_{2})^{2}

, you lose the essential conjugate terms that allow for the beautiful simplification we see here.

Why does the conjugate of

a

remain

a

in the expansion?

Great observation! Remember that

a

and

b

are given as real numbers. For any real number

x

, the conjugate

\overset{x}{ˉ}

is simply

x

because the imaginary part is zero. This property is what allows us to treat

a

and

b

as constants during the conjugation process, making the algebra much cleaner.

Is this result just a coincidence, or is there a deeper meaning?

It is definitely not a coincidence! This identity is a beautiful, generalized version of the Parallelogram Law for complex numbers. When you set

a = 1

and

b = 1

, the expression simplifies to

2 (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2})

, which is the standard form of the Parallelogram Law. You have just derived a powerful generalization!

Animated Solution for Mathematics - Complex Numbers: For any two complex numbers $z_{1}, z_{2}$ and any real number $a$ and $b$ , $∣ a z_{1} - b z_{2} ∣^{2} + ∣ b z_{1} + a z_{2} ∣^{2} = \dots$

Visualized Solution

Introduction to the Expression

The Modulus-Conjugate Tool

Applying the Property to Term 1

Applying the Property to Term 2

Expanding the First Product

Expanding the Second Product

Combining and Cancelling

Grouping and Factorizing

Final Takeaway

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Master Key

The Algebraic Dance

The Moment of Cancellation

Final Synthesis

Similar Questions

Sum of squares of modulus of all the complex numbers $z$ satisfying $\overset{z}{ˉ} = i z^{2} + z^{2} - z$ is equal to \_\_\_\_\_.

Let $z_{1}$ and $z_{2}$ be two complex number such that $z_{1} + z_{2} = 5$ and $z_{1}^{3} + z_{2}^{3} = 20 + 15 i$ . Then $∣ z_{1}^{4} + z_{2}^{4} ∣$ equals

For complex numbers $z$ and $w$ , prove that $∣ z ∣^{2} w - ∣ w ∣^{2} z = z - w$ if and only if $z = w$ or $z \overset{w}{ˉ} = 1$ .

Let $ω = e^{iπ /3}$ , and $a, b, c, x, y, z$ be non-zero complex numbers such that $a + b + c = x, a + bω + c ω^{2} = y, a + b ω^{2} + c ω = z$ . Then the value of $\frac{∣ x ∣ ^{2} + ∣ y ∣ ^{2} + ∣ z ∣ ^{2}}{∣ a ∣ ^{2} + ∣ b ∣ ^{2} + ∣ c ∣ ^{2}}$ is .........

If $z = x - i y$ and $z^{1/3} = p + i q$ , then $(x / p + y / q) / (p^{2} + q^{2})$ is equal to

Let $z$ be a complex number with non-zero imaginary part. If $\frac{2 + 3 z + 4 z ^{2}}{2 - 3 z + 4 z ^{2}}$ is a real number, then the value of $∣ z ∣^{2}$ is _____________.

If $z = \frac{1}{2} - 2 i$ , is such that $∣ z + 1∣ = α z + β (1 + i), i = - 1$ and $α, β \in R$ , then $α + β$ is equal to

Let $S = {z \in C : z^{2} + \overset{z}{ˉ} = 0}$ . Then $\sum_{z \in S} (Re (z) + Im (z))$ 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; } For any two complex numbers z1​,z2​ and any real number a and b, ∣az1​−bz2​∣2+∣bz1​+az2​∣2=…

Visualized Solution

Introduction to the Expression

The Modulus-Conjugate Tool

Applying the Property to Term 1

Applying the Property to Term 2

Expanding the First Product

Expanding the Second Product

Combining and Cancelling

Grouping and Factorizing

Final Takeaway

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Master Key

The Algebraic Dance

The Moment of Cancellation

Final Synthesis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sum of squares of modulus of all the complex numbers z satisfying zˉ=iz2+z2−z 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 complex number such that z1​+z2​=5 and z13​+z23​=20+15i. Then ∣z14​+z24​∣ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For complex numbers z and w, prove that ∣z∣2w−∣w∣2z=z−w if and only if z=w or zwˉ=1.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=x−iy and z1/3=p+iq, then (x/p+y/q)/(p2+q2) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be a complex number with non-zero imaginary part. If 2−3z+4z22+3z+4z2​ is a real number, then the value of ∣z∣2 is _____________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=21​−2i, is such that ∣z+1∣=αz+β(1+i),i=−1​ and α,β∈R, 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 S={z∈C:z2+zˉ=0}. Then ∑z∈S​(Re(z)+Im(z)) is equal to

Animated Solution for Mathematics - Complex Numbers: For any two complex numbers $z_{1}, z_{2}$ and any real number $a$ and $b$ , $∣ a z_{1} - b z_{2} ∣^{2} + ∣ b z_{1} + a z_{2} ∣^{2} = \dots$

Sum of squares of modulus of all the complex numbers $z$ satisfying $\overset{z}{ˉ} = i z^{2} + z^{2} - z$ is equal to \_\_\_\_\_.

Let $z_{1}$ and $z_{2}$ be two complex number such that $z_{1} + z_{2} = 5$ and $z_{1}^{3} + z_{2}^{3} = 20 + 15 i$ . Then $∣ z_{1}^{4} + z_{2}^{4} ∣$ equals

For complex numbers $z$ and $w$ , prove that $∣ z ∣^{2} w - ∣ w ∣^{2} z = z - w$ if and only if $z = w$ or $z \overset{w}{ˉ} = 1$ .

If $z = x - i y$ and $z^{1/3} = p + i q$ , then $(x / p + y / q) / (p^{2} + q^{2})$ is equal to

Let $z$ be a complex number with non-zero imaginary part. If $\frac{2 + 3 z + 4 z ^{2}}{2 - 3 z + 4 z ^{2}}$ is a real number, then the value of $∣ z ∣^{2}$ is _____________.

If $z = \frac{1}{2} - 2 i$ , is such that $∣ z + 1∣ = α z + β (1 + i), i = - 1$ and $α, β \in R$ , then $α + β$ is equal to

Let $S = {z \in C : z^{2} + \overset{z}{ˉ} = 0}$ . Then $\sum_{z \in S} (Re (z) + Im (z))$ is equal to