Let A = {θ ∈ (0, 2π) : (1 + 2isinθ)/(1 - isinθ) is purely imaginary}. Then the sum of the elements i

Q: Why do we multiply by the conjugate of the denominator?

When dealing with complex numbers in the form $\frac{z_1}{z_2}$, we cannot easily see the real and imaginary parts. By multiplying by the conjugate $\bar{z_2}$, we make the denominator a real number, which is the standard procedure for simplifying complex fractions.

Q: Do we need to worry about the denominator being zero?

A great habit to have! In this problem, the denominator is $1 + \sin^2\theta$. Since $\sin^2\theta \ge 0$, the denominator is at least 1, so it is never zero. We are safe.

Q: Why is the range for $2\theta$ from $0$ to $4\pi$?

Since $\theta$ is in $(0, 2\pi)$, multiplying by 2 gives $0 < 2\theta < 4\pi$. This covers two full rotations, which is necessary to capture all possible solutions for $\cos 2\theta = 0$.

Question

Let $A = \{	heta \in (0, 2\pi) : \frac{1 + 2i\sin	heta}{1 - i\sin	heta} 	ext{ is purely imaginary}\}$. Then the sum of the elements in $A$ is

Options:

(A) $4\pi$

(B) $3\pi$

(C) $\pi$

(D) $2\pi$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Argand plane, the complex playground where real numbers live on the horizontal axis and imaginary numbers dance on the vertical axis. Our problem asks us to find values of $	heta$ such that the complex number $z$ is purely imaginary:

$$z = \frac{1 + 2i\sin	heta}{1 - i\sin	heta}$$

For $z$ to be purely imaginary, it must lie entirely on the vertical axis. This is a fancy way of saying its **real part must be zero**.

### The Art of Rationalization

When you see a complex number in the denominator, your first instinct should always be to rationalize. We multiply both the numerator and the denominator by the conjugate of the denominator, which is $1 + i\sin	heta$.

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

The denominator simplifies beautifully to $1^2 - (i\sin	heta)^2 = 1 + \sin^2	heta$. Now, the denominator is a purely real number, which is exactly what we wanted.

### The Extraction of the Real Part

Now, let us expand the numerator:

$$(1 + 2i\sin	heta)(1 + i\sin	heta) = 1 + i\sin	heta + 2i\sin	heta + 2i^2\sin^2	heta$$

Since $i^2 = -1$, this simplifies to $1 + 3i\sin	heta - 2\sin^2	heta$. Grouping the real and imaginary parts, we get:

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

We can now clearly see the real part:

$$	ext{Re}(z) = \frac{1 - 2\sin^2	heta}{1 + \sin^2	heta}$$

For $z$ to be purely imaginary, we set this real part to zero, which implies $1 - 2\sin^2	heta = 0$.

### The Trigonometric Dance

This equation, $1 - 2\sin^2	heta = 0$, is a classic. It is the double-angle identity for cosine in disguise, as $\cos 2	heta = 1 - 2\sin^2	heta$.

Thus, our condition simplifies to $\cos 2	heta = 0$. Given $	heta \in (0, 2\pi)$, it follows that $2	heta \in (0, 4\pi)$.

We need to find all values of $2	heta$ in this range where the cosine function vanishes. These are the odd multiples of $\frac{\pi}{2}$:

$$2	heta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$$

Dividing by 2, we find our four values for $	heta$:

$$	heta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$

### Final Calculation

We have found our four angles. The final step is to sum them up:

$$\frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi$$

The final result is **$4\pi$**. It is elegant, it is precise, and it is the beauty of complex numbers.

Animated Solution for Mathematics - Complex Numbers: 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

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Visualized Solution

$z = \frac{1 + 2 i s i n θ}{1 - i s i n θ}$

Rationalizing the Denominator

Setting up the Multiplication

Simplifying the Denominator

Expanding the Numerator

Grouping Real and Imaginary Parts

Applying the Purely Imaginary Condition

Solving the Trig Equation

Determining the Range for $2 θ$

Finding Values of $2 θ$

Solving for $θ$

Calculating the Sum

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Art of Rationalization

The Extraction of the Real Part

The Trigonometric Dance

Final Calculation

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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−isinθ1+2isinθ​ is purely imaginary}. Then the sum of the elements in A is