Let A = { θ ∈ (-π/2, π) : (3 + 2i sin θ) / (1 - 2i sin θ) is purely imaginary }. Then the sum of the

Q: Wait, why do we need to rationalize? Can't we just set the real part of the numerator to zero?

That is a common trap! A complex number $z = \frac{N}{D}$ is purely imaginary if $\text{Re}(z) = 0$. If you only look at the numerator, you are ignoring the denominator's contribution to the real part. Rationalizing ensures the denominator becomes a real number, allowing you to clearly separate the real and imaginary components of the entire fraction.

Q: How do I know which values of $\theta$ to keep in the interval $(-\frac{\pi}{2}, \pi)$?

Great question. The interval $(-\frac{\pi}{2}, \pi)$ spans from the fourth quadrant ($-90^\circ$) through the first and second quadrants up to $180^\circ$. When you solve $\sin \theta = \pm \frac{\sqrt{3}}{2}$, you must check each solution against this range. For example, $\sin \theta = \frac{\sqrt{3}}{2}$ gives $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, both of which are in the interval. $\sin \theta = -\frac{\sqrt{3}}{2}$ gives $-\frac{\pi}{3}$, which is also in the interval.

Q: Why is the denominator $1 + 4 \sin^2 \theta$ always real?

Because it is the product of a complex number and its conjugate: $(1 - 2i \sin \theta)(1 + 2i \sin \theta)$. Using the identity $(a - bi)(a + bi) = a^2 + b^2$, we get $1^2 + (2 \sin \theta)^2 = 1 + 4 \sin^2 \theta$. Since $\sin^2 \theta$ is always non-negative, this denominator is always positive and real, which is exactly what we need to safely set the real part of the numerator to zero.

Question

Let $A = \{ 	heta \in (-\pi/2, \pi) : \frac{3 + 2i \sin 	heta}{1 - 2i \sin 	heta} 	ext{ is purely imaginary} \}$. Then the sum of the elements in A is :

Options:

(A) $\frac{5\pi}{6}$

(B) $\frac{2\pi}{3}$

(C) $\frac{3\pi}{4}$

(D) $\pi$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the complex plane. You have a complex number $z = \frac{3 + 2i \sin 	heta}{1 - 2i \sin 	heta}$.

The problem states that this number is **purely imaginary**. This means that if you were to plot this number on the Argand plane, it would lie perfectly on the vertical axis, with no horizontal displacement.

In other words, its real part is zero. This is our guiding light; we only need to force the real part to vanish.

### The Art of Rationalization

Currently, our expression is a messy fraction. We have an imaginary term in the denominator, which makes it impossible to see the real and imaginary parts clearly.

To fix this, we use the most powerful tool in our complex number toolkit: the conjugate. We multiply both the numerator and the denominator by $1 + 2i \sin 	heta$.

The denominator becomes $(1 - 2i \sin 	heta)(1 + 2i \sin 	heta)$, which simplifies using the identity $(a - b)(a + b) = a^2 - b^2$. Since $i^2 = -1$, the denominator transforms into:

$$1 + 4 \sin^2 	heta$$

It is now a purely real, positive number.

### The Algebraic Expansion

Now, let us turn our attention to the numerator. We multiply $(3 + 2i \sin 	heta)(1 + 2i \sin 	heta)$.

Expanding this term by term, we get $3(1) + 3(2i \sin 	heta) + (2i \sin 	heta)(1) + (2i \sin 	heta)(2i \sin 	heta)$. This simplifies to $3 + 6i \sin 	heta + 2i \sin 	heta + 4i^2 \sin^2 	heta$.

Since $i^2 = -1$, the last term becomes $-4 \sin^2 	heta$. Grouping the real and imaginary parts, we have $(3 - 4 \sin^2 	heta) + i(8 \sin 	heta)$.

Now, our entire complex number $z$ is written as:

$$z = \frac{(3 - 4 \sin^2 	heta) + i(8 \sin 	heta)}{1 + 4 \sin^2 	heta}$$

### The Moment of Truth

We know that for $z$ to be purely imaginary, its real part must be zero. Therefore, we set the real part of our fraction to zero:

$$\frac{3 - 4 \sin^2 	heta}{1 + 4 \sin^2 	heta} = 0$$

A fraction is zero only when its numerator is zero. Thus, $3 - 4 \sin^2 	heta = 0$, which leads us to:

$$\sin^2 	heta = \frac{3}{4}$$

Taking the square root, we find $\sin 	heta = \pm \frac{\sqrt{3}}{2}$.

### The Unit Circle Journey

We are restricted to the interval $	heta \in (-\frac{\pi}{2}, \pi)$. Let us map these solutions.

For $\sin 	heta = \frac{\sqrt{3}}{2}$, we look at the first and second quadrants. The solutions are $	heta = \frac{\pi}{3}$ and $	heta = \frac{2\pi}{3}$.

For $\sin 	heta = -\frac{\sqrt{3}}{2}$, we look at the fourth quadrant, giving us $	heta = -\frac{\pi}{3}$. All three values fall within our allowed interval.

Finally, we sum these elements: $\frac{\pi}{3} + \frac{2\pi}{3} - \frac{\pi}{3}$. The $\frac{\pi}{3}$ and $-\frac{\pi}{3}$ cancel out perfectly.

The final result is:

$$\mathbf{\frac{2\pi}{3}}$$

Animated Solution for Mathematics - Complex Numbers: 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 :

Select Answer:

Visualized Solution

Condition for Purely Imaginary

Rationalizing the Denominator

Expanding the Denominator

Expanding the Numerator

Separating Real and Imaginary Parts

Equating Real Part to Zero

Solving for $sin θ$

Visualizing the Given Interval

Solutions for $sin θ = \frac{3}{2}$

Solutions for $sin θ = - \frac{3}{2}$

Summing the Elements of Set A

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Art of Rationalization

The Algebraic Expansion

The Moment of Truth

The Unit Circle Journey

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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∈(−π/2,π):1−2isinθ3+2isinθ​ is purely imaginary}. Then the sum of the elements in A is :