Let $A = \{\theta \in [0, 2\pi] : 1 + 10 \text{ Re}(\frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta}) = 0\}$. Then $\sum_{\theta \in A} \theta^2$ is equal to Options: (A) $\frac{21}{4}\pi^2$ (B) $8\pi^2$ (C) $\frac{27}{4}\pi^2$ (D) $6\pi^2$

### Analyzing the Setup We are given the complex number $z$ defined by the expression: $$z = \frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta}$$ To find the real part, $\text{Re}(z)$, we employ the **conjugate trick**. We multiply both the numerator and the denominator by the complex conjugate of the denominator, which is $\cos\theta + 3i\sin\theta$. ### The Master Equation Multiplying the numerator and denominator yields: $$z = \frac{(2\cos\theta + i\sin\theta)(\cos\theta + 3i\sin\theta)}{\cos^2\theta + 9\sin^2\theta}$$ Expanding the numerator, we obtain $2\cos^2\theta + 6i\sin\theta\cos\theta + i\sin\theta\cos\theta + 3i^2\sin^2\theta$. Recalling that $i^2 = -1$, the real part of the numerator simplifies to $2\cos^2\theta - 3\sin^2\theta$. Thus, the real part of $z$ is: $$\text{Re}(z) = \frac{2\cos^2\theta - 3\sin^2\theta}{\cos^2\theta + 9\sin^2\theta}$$ ### Solving the Condition We are given the condition $1 + 10 \text{Re}(z) = 0$. Substituting our expression for $\text{Re}(z)$, we get: $$1 + 10 \left( \frac{2\cos^2\theta - 3\sin^2\theta}{\cos^2\theta + 9\sin^2\theta} \right) = 0$$ Multiplying through by the denominator, we obtain: $$(\cos^2\theta + 9\sin^2\theta) + 10(2\cos^2\theta - 3\sin^2\theta) = 0$$ Simplifying the terms leads to: $$21\cos^2\theta - 21\sin^2\theta = 0 \Rightarrow \tan^2\theta = 1$$ ### Final Calculation In the interval $[0, 2\pi]$, the solutions for $\tan\theta = \pm 1$ are $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \text{and } \frac{7\pi}{4}$. We are required to find the sum of the squares of these values: $$\text{Sum} = \left( \frac{\pi}{4} \right)^2 + \left( \frac{3\pi}{4} \right)^2 + \left( \frac{5\pi}{4} \right)^2 + \left( \frac{7\pi}{4} \right)^2$$ $$\text{Sum} = \frac{\pi^2}{16} (1 + 9 + 25 + 49) = \frac{84\pi^2}{16}$$ The final result is: $$\mathbf{\frac{21}{4}\pi^2}$$

Wait, why does the denominator become $\cos^2\theta + 9\sin^2\theta$ and not $\cos^2\theta - 9\sin^2\theta$?

Ah, the classic $i^2$ trap! Remember, the denominator is $(\cos\theta - 3i\sin\theta)(\cos\theta + 3i\sin\theta)$. This is a difference of squares: $a^2 - (3i\sin\theta)^2$. Since $i^2 = -1$, the minus sign becomes a plus. Always watch out for that $i^2$ sign flip!

Is it possible that the denominator is zero for some $\theta$?

Excellent mathematical intuition! We must always check the domain. If $\cos^2\theta + 9\sin^2\theta = 0$, the expression is undefined. Since $\cos^2\theta$ and $\sin^2\theta$ are non-negative, this only happens if both are zero, which is impossible. So, our solutions are safe!

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Animated Solution for Mathematics - Complex Numbers: Let $A = {θ \in [0, 2 π] : 1 + 10 Re (\frac{2 c o s θ + i s i n θ}{c o s θ - 3 i s i n θ}) = 0}$ . Then $\sum_{θ \in A} θ^{2}$ is equal to

Select Answer:

Visualized Solution

Defining the Complex Expression $z$

Let $z = \frac{2 c o s θ + i s i n θ}{c o s θ - 3 i s i n θ}$
The given condition is $1 + 10 Re (z) = 0$

Rationalizing the Denominator of $z$

To find $Re (z)$ , multiply numerator and denominator by the conjugate of the denominator:
$z = \frac{( 2 c o s θ + i s i n θ ) ( c o s θ + 3 i s i n θ )}{( c o s θ - 3 i s i n θ ) ( c o s θ + 3 i s i n θ )}$

Expanding the Numerator

Numerator expansion:
$(2 cos θ) (cos θ) + (2 cos θ) (3 i sin θ) + (i sin θ) (cos θ) + (i sin θ) (3 i sin θ)$
$= 2 cos^{2} θ + 7 i sin θ cos θ + 3 i^{2} sin^{2} θ$
Using $i^{2} = - 1$ :
Numerator $= (2 cos^{2} θ - 3 sin^{2} θ) + i (7 sin θ cos θ)$

Extracting $Re (z)$

Denominator expansion:
$(cos θ)^{2} - (3 i sin θ)^{2} = cos^{2} θ - 9 i^{2} sin^{2} θ$
$= cos^{2} θ + 9 sin^{2} θ$
Therefore, $Re (z) = \frac{2 c o s ^{2} θ - 3 s i n ^{2} θ}{c o s ^{2} θ + 9 s i n ^{2} θ}$

Substituting into $1 + 10 Re (z) = 0$

Substitute $Re (z)$ into $1 + 10 Re (z) = 0$ :
$1 + 10 (\frac{2 c o s ^{2} θ - 3 s i n ^{2} θ}{c o s ^{2} θ + 9 s i n ^{2} θ}) = 0$
Multiply through by $(cos^{2} θ + 9 sin^{2} θ)$ :
$(cos^{2} θ + 9 sin^{2} θ) + 10 (2 cos^{2} θ - 3 sin^{2} θ) = 0$

Solving for $tan^{2} θ$

Expand and group terms:
$cos^{2} θ + 20 cos^{2} θ + 9 sin^{2} θ - 30 sin^{2} θ = 0$
$21 cos^{2} θ - 21 sin^{2} θ = 0$
$cos^{2} θ = sin^{2} θ$
$tan^{2} θ = 1 ⟹ tan θ = \pm 1$

Finding Solutions in $[0, 2 π]$

For $θ \in [0, 2 π]$ , we need angles where $tan θ = 1$ or $tan θ = - 1$ .
In the first quadrant: $θ = \frac{π}{4}$
In the second quadrant: $θ = \frac{3 π}{4}$
In the third quadrant: $θ = \frac{5 π}{4}$
In the fourth quadrant: $θ = \frac{7 π}{4}$

Squaring the Solutions $θ^{2}$

Calculate $θ^{2}$ for each value:
$(\frac{π}{4})^{2} = \frac{π ^{2}}{16}$
$(\frac{3 π}{4})^{2} = \frac{9 π ^{2}}{16}$
$(\frac{5 π}{4})^{2} = \frac{25 π ^{2}}{16}$
$(\frac{7 π}{4})^{2} = \frac{49 π ^{2}}{16}$

Summing the Squares $\sum θ^{2}$

Sum $= \frac{π ^{2}}{16} (1 + 9 + 25 + 49)$
Sum $= \frac{84 π ^{2}}{16}$
Simplify the fraction: $\frac{84}{16} = \frac{21}{4}$
Final Sum $= \frac{21}{4} π^{2}$

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The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Analyzing the Setup

We are given the complex number

z

defined by the expression:

z = \frac{2 cos θ + i sin θ}{cos θ - 3 i sin θ}

To find the real part,

Re (z)

, we employ the conjugate trick. We multiply both the numerator and the denominator by the complex conjugate of the denominator, which is

cos θ + 3 i sin θ

The Master Equation

Multiplying the numerator and denominator yields:

z = \frac{( 2 cos θ + i sin θ ) ( cos θ + 3 i sin θ )}{cos ^{2} θ + 9 sin ^{2} θ}

Expanding the numerator, we obtain

2 cos^{2} θ + 6 i sin θ cos θ + i sin θ cos θ + 3 i^{2} sin^{2} θ

. Recalling that

i^{2} = - 1

, the real part of the numerator simplifies to

2 cos^{2} θ - 3 sin^{2} θ

Thus, the real part of

z

is:

Re (z) = \frac{2 cos ^{2} θ - 3 sin ^{2} θ}{cos ^{2} θ + 9 sin ^{2} θ}

Solving the Condition

We are given the condition

1 + 10 Re (z) = 0

. Substituting our expression for

Re (z)

, we get:

1 + 10 (\frac{2 cos ^{2} θ - 3 sin ^{2} θ}{cos ^{2} θ + 9 sin ^{2} θ}) = 0

Multiplying through by the denominator, we obtain:

(cos^{2} θ + 9 sin^{2} θ) + 10 (2 cos^{2} θ - 3 sin^{2} θ) = 0

Simplifying the terms leads to:

21 cos^{2} θ - 21 sin^{2} θ = 0 \Rightarrow tan^{2} θ = 1

Final Calculation

In the interval

[0, 2 π]

, the solutions for

tan θ = \pm 1

are

θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, and \frac{7 π}{4}

We are required to find the sum of the squares of these values:

Sum = (\frac{π}{4})^{2} + (\frac{3 π}{4})^{2} + (\frac{5 π}{4})^{2} + (\frac{7 π}{4})^{2}

Sum = \frac{π ^{2}}{16} (1 + 9 + 25 + 49) = \frac{84 π ^{2}}{16}

The final result is:

\frac{21}{4} π^{2}

Common Traps & FAQs

Why do we multiply the numerator and denominator by the conjugate? Can't we just separate the real and imaginary parts directly?

Great question! You cannot separate them directly because the

i

is trapped in the denominator. Think of the conjugate as a 'key' that unlocks the denominator, turning it into a real number so you can finally see the real and imaginary parts of the numerator clearly.

Wait, why does the denominator become

cos^{2} θ + 9 sin^{2} θ

and not

cos^{2} θ - 9 sin^{2} θ

Ah, the classic

i^{2}

trap! Remember, the denominator is

(cos θ - 3 i sin θ) (cos θ + 3 i sin θ)

. This is a difference of squares:

a^{2} - (3 i sin θ)^{2}

. Since

i^{2} = - 1

, the minus sign becomes a plus. Always watch out for that

i^{2}

sign flip!

Is it possible that the denominator is zero for some

θ

Excellent mathematical intuition! We must always check the domain. If

cos^{2} θ + 9 sin^{2} θ = 0

, the expression is undefined. Since

cos^{2} θ

and

sin^{2} θ

are non-negative, this only happens if both are zero, which is impossible. So, our solutions are safe!

Animated Solution for Mathematics - Complex Numbers: Let $A = {θ \in [0, 2 π] : 1 + 10 Re (\frac{2 c o s θ + i s i n θ}{c o s θ - 3 i s i n θ}) = 0}$ . Then $\sum_{θ \in A} θ^{2}$ is equal to

Select Answer:

Visualized Solution

Defining the Complex Expression $z$

Rationalizing the Denominator of $z$

Expanding the Numerator

Extracting $Re (z)$

Substituting into $1 + 10 Re (z) = 0$

Solving for $tan^{2} θ$

Finding Solutions in $[0, 2 π]$

Squaring the Solutions $θ^{2}$

Summing the Squares $\sum θ^{2}$

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Master Equation

Solving the Condition

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={θ∈[0,2π]:1+10 Re(cosθ−3isinθ2cosθ+isinθ​)=0}. Then ∑θ∈A​θ2 is equal to

Select Answer:

Visualized Solution

Defining the Complex Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Rationalizing the Denominator of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Expanding the Numerator

Extracting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Re(z)

Substituting into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+10Re(z)=0

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan2θ

Finding Solutions in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,2π]

Squaring the Solutions .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ2

Summing the Squares .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑θ2

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Master Equation

Solving the Condition

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Complex Numbers: Let $A = {θ \in [0, 2 π] : 1 + 10 Re (\frac{2 c o s θ + i s i n θ}{c o s θ - 3 i s i n θ}) = 0}$ . Then $\sum_{θ \in A} θ^{2}$ is equal to

Defining the Complex Expression $z$

Rationalizing the Denominator of $z$

Extracting $Re (z)$

Substituting into $1 + 10 Re (z) = 0$

Solving for $tan^{2} θ$

Finding Solutions in $[0, 2 π]$

Squaring the Solutions $θ^{2}$

Summing the Squares $\sum θ^{2}$

Let $A = {θ \in (0, 2 π) : \frac{1 + 2 i s i n θ}{1 - i s i n θ} is purely imaginary}$ . Then the sum of the elements in $A$ is

Let $A = {θ \in (- π /2, π) : \frac{3 + 2 i s i n θ}{1 - 2 i s i n θ} is purely imaginary}$ . Then the sum of the elements in A is :

The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i c o s θ}{1 - 2 i c o s θ}$ is purely imaginary, is equal

Let $S = {z \in C : \overset{z}{ˉ} = i (z^{2} + Re (\overset{z}{ˉ}))}$ . Then $\sum_{z \in S} ∣ z ∣^{2}$ 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

Let $z \in C$ be such that $\frac{z ^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Then the sum of all possible values of $z^{2}$ is

Let $n$ denote the number of solutions of the equation $z^{2} + 3 \overset{z}{ˉ} = 0$ , where $z$ is a complex number. Then the value of $\sum_{k = 0}^{\infty} \frac{1}{n ^{k}}$ is equal to

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