Let the product of $\omega_1 = (8 + i)\sin\theta + (7 + 4i)\cos\theta$ and $\omega_2 = (1 + 8i)\sin\theta + (4 + 7i)\cos\theta$ be $\alpha + i\beta$, $i = \sqrt{-1}$. Let $p$ and $q$ be the maximum and the minimum values of $\alpha + \beta$ respectively. Then $p + q$ is equal to Options: (A) 140 (B) 130 (C) 160 (D) 150

### Analyzing the Setup We are given two complex numbers: $\omega_1 = (8 + i)\sin\theta + (7 + 4i)\cos\theta$ $\omega_2 = (1 + 8i)\sin\theta + (4 + 7i)\cos\theta$ By separating the real and imaginary components, we observe: $\omega_1 = (8\sin\theta + 7\cos\theta) + i(\sin\theta + 4\cos\theta)$ $\omega_2 = (\sin\theta + 4\cos\theta) + i(8\sin\theta + 7\cos\theta)$ ### The Master Equation Let us define the variables $x$ and $y$ as follows: $x = 8\sin\theta + 7\cos\theta$ $y = \sin\theta + 4\cos\theta$ Substituting these into our expressions, we find $\omega_1 = x + iy$ and $\omega_2 = y + ix$. Multiplying these two complex numbers yields: $$\omega_1 \omega_2 = (x + iy)(y + ix) = xy + ix^2 + iy^2 + i^2xy$$ Since $i^2 = -1$, the expression simplifies to: $$\omega_1 \omega_2 = i(x^2 + y^2)$$ Here, the real part $\alpha$ is $0$, and the imaginary part $\beta$ is $x^2 + y^2$. Our goal is to determine the range of $x^2 + y^2$. ### Final Calculation Expanding the sum of squares, we have: $$x^2 + y^2 = (8\sin\theta + 7\cos\theta)^2 + (\sin\theta + 4\cos\theta)^2$$ Expanding the squares: $$x^2 + y^2 = (64\sin^2\theta + 49\cos^2\theta + 112\sin\theta\cos\theta) + (\sin^2\theta + 16\cos^2\theta + 8\sin\theta\cos\theta)$$ Combining like terms: $$x^2 + y^2 = 65\sin^2\theta + 65\cos^2\theta + 120\sin\theta\cos\theta$$ $$x^2 + y^2 = 65 + 60\sin(2\theta)$$ The range of $\sin(2\theta)$ is $[-1, 1]$. Therefore, the maximum value is $65 + 60 = 125$, and the minimum value is $65 - 60 = 5$. The sum of the maximum and minimum values is: **$125 + 5 = 130$**

Why did the $xy$ terms cancel out in the product?

When you multiply $(x+iy)(y+ix)$, you get $xy + ix^2 + iy^2 + i^2xy$. Since $i^2 = -1$, the $xy$ and $-xy$ terms cancel out, leaving you with $i(x^2+y^2)$. It is a beautiful example of how structure simplifies complexity.

Can I just multiply the complex numbers directly without substitution?

You certainly can, but it is a path filled with algebraic traps. The substitution method is not just faster; it is safer and reveals the underlying symmetry of the problem, which is a hallmark of JEE Advanced questions.

How do I know $\alpha=0$?

The problem defines the product as $\alpha + i\beta$. After simplifying the product to $i(x^2+y^2)$, we see the real part is $0$. Thus, $\alpha$ must be $0$, and $\beta$ is $x^2+y^2$.

JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Let the product of $ω_{1} = (8 + i) sin θ + (7 + 4 i) cos θ$ and $ω_{2} = (1 + 8 i) sin θ + (4 + 7 i) cos θ$ be $α + i β$ , $i = - 1$ . Let $p$ and $q$ be the maximum and the minimum values of $α + β$ respectively. Then $p + q$ is equal to

Select Answer:

Visualized Solution

Analyze $ω_{1}$

$ω_{1} = (8 + i) sin θ + (7 + 4 i) cos θ$
Rearranging real and imaginary parts:
$ω_{1} = (8 sin θ + 7 cos θ) + i (sin θ + 4 cos θ)$

Analyze $ω_{2}$

$ω_{2} = (1 + 8 i) sin θ + (4 + 7 i) cos θ$
Rearranging real and imaginary parts:
$ω_{2} = (sin θ + 4 cos θ) + i (8 sin θ + 7 cos θ)$

Substitution for Simplicity

Let $x = 8 sin θ + 7 cos θ$
Let $y = sin θ + 4 cos θ$
Then, $ω_{1} = x + i y$
And $ω_{2} = y + i x$

The Product $ω_{1} ω_{2}$

$ω_{1} ω_{2} = (x + i y) (y + i x)$
$= x y + i x^{2} + i y^{2} + i^{2} x y$
Since $i^{2} = - 1$ :
$= x y + i (x^{2} + y^{2}) - x y$
$= i (x^{2} + y^{2})$

Identify $α$ and $β$

Given: $ω_{1} ω_{2} = α + i β$
We found: $ω_{1} ω_{2} = 0 + i (x^{2} + y^{2})$
Comparing real and imaginary parts:
$α = 0$
$β = x^{2} + y^{2}$
Therefore, $α + β = x^{2} + y^{2}$

Expand $x^{2} + y^{2}$

$α + β = (8 sin θ + 7 cos θ)^{2} + (sin θ + 4 cos θ)^{2}$
Expanding the squares:
$= (64 sin^{2} θ + 49 cos^{2} θ + 112 sin θ cos θ) + (sin^{2} θ + 16 cos^{2} θ + 8 sin θ cos θ)$

Simplify the Expression

Grouping like terms:
$= (64 + 1) sin^{2} θ + (49 + 16) cos^{2} θ + (112 + 8) sin θ cos θ$
$= 65 sin^{2} θ + 65 cos^{2} θ + 120 sin θ cos θ$
$= 65 (sin^{2} θ + cos^{2} θ) + 120 sin θ cos θ$
$= 65 + 120 sin θ cos θ$

Use Double Angle Identity

Recall: $2 sin θ cos θ = sin 2 θ$
Rewrite the expression:
$α + β = 65 + 60 (2 sin θ cos θ)$
$α + β = 65 + 60 sin 2 θ$

Find Maximum Value ( $p$ )

The maximum value of $sin 2 θ$ is $1$ .
Maximum value $p = 65 + 60 (1)$
$p = 125$

Find Minimum Value ( $q$ )

The minimum value of $sin 2 θ$ is $- 1$ .
Minimum value $q = 65 + 60 (- 1)$
$q = 65 - 60 = 5$

Final Calculation

We need to find $p + q$ .
$p + q = 125 + 5$
$p + q = 130$
The final answer is 130.

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Analyzing the Setup

We are given two complex numbers:

ω_{1} = (8 + i) sin θ + (7 + 4 i) cos θ

ω_{2} = (1 + 8 i) sin θ + (4 + 7 i) cos θ

By separating the real and imaginary components, we observe:

ω_{1} = (8 sin θ + 7 cos θ) + i (sin θ + 4 cos θ)

ω_{2} = (sin θ + 4 cos θ) + i (8 sin θ + 7 cos θ)

The Master Equation

Let us define the variables

x

and

y

as follows:

x = 8 sin θ + 7 cos θ

y = sin θ + 4 cos θ

Substituting these into our expressions, we find

ω_{1} = x + i y

and

ω_{2} = y + i x

. Multiplying these two complex numbers yields:

ω_{1} ω_{2} = (x + i y) (y + i x) = x y + i x^{2} + i y^{2} + i^{2} x y

Since

i^{2} = - 1

, the expression simplifies to:

ω_{1} ω_{2} = i (x^{2} + y^{2})

Here, the real part

α

0

, and the imaginary part

β

x^{2} + y^{2}

. Our goal is to determine the range of

x^{2} + y^{2}

Final Calculation

Expanding the sum of squares, we have:

x^{2} + y^{2} = (8 sin θ + 7 cos θ)^{2} + (sin θ + 4 cos θ)^{2}

Expanding the squares:

x^{2} + y^{2} = (64 sin^{2} θ + 49 cos^{2} θ + 112 sin θ cos θ) + (sin^{2} θ + 16 cos^{2} θ + 8 sin θ cos θ)

Combining like terms:

x^{2} + y^{2} = 65 sin^{2} θ + 65 cos^{2} θ + 120 sin θ cos θ

x^{2} + y^{2} = 65 + 60 sin (2 θ)

The range of

sin (2 θ)

[- 1, 1]

. Therefore, the maximum value is

65 + 60 = 125

, and the minimum value is

65 - 60 = 5

The sum of the maximum and minimum values is: $125 + 5 = 130$

Select Answer:

Visualized Solution

Analyze $ω_{1}$

Analyze $ω_{2}$

Substitution for Simplicity

The Product $ω_{1} ω_{2}$

Identify $α$ and $β$

Expand $x^{2} + y^{2}$

Simplify the Expression

Use Double Angle Identity

Find Maximum Value ( $p$ )

Find Minimum Value ( $q$ )

Final Calculation

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

Analyze .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω1​

Analyze .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω2​

Substitution for Simplicity

The Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω1​ω2​

Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Expand .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+y2

Simplify the Expression

Use Double Angle Identity

Find Maximum Value ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p)

Find Minimum Value ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q)

Final Calculation

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Master Equation

Final Calculation

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Find Maximum Value ( $p$ )

Find Minimum Value ( $q$ )

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