A value of $\theta$ for which $\frac{2 + 3i \sin \theta}{1 - 2i \sin \theta}$ is purely imaginary, is Options: (A) $\sin^{-1}(\sqrt{3}/4)$ (B) $\sin^{-1}(1/\sqrt{3})$ (C) $\pi/3$ (D) $\pi/6$

### Understanding the Purely Imaginary Condition In the Argand plane, a complex number $Z = x + iy$ represents a point where $x$ is the real part and $y$ is the imaginary part. When we define a number as **purely imaginary**, we are stating that it has no horizontal component. Geometrically, this means the real part must vanish, so we set $x = 0$. ### Untangling the Expression We are given the complex expression: $$Z = \frac{2 + 3i \sin \theta}{1 - 2i \sin \theta}$$ To simplify this, we employ the technique of **rationalization**. We multiply both the numerator and the denominator by the complex conjugate of the denominator, which is $1 + 2i \sin \theta$. The denominator becomes: $$(1 - 2i \sin \theta)(1 + 2i \sin \theta) = 1^2 - (2i \sin \theta)^2 = 1 + 4 \sin^2 \theta$$ ### Expanding the Numerator Next, we expand the numerator: $$(2 + 3i \sin \theta)(1 + 2i \sin \theta) = 2 + 4i \sin \theta + 3i \sin \theta + 6i^2 \sin^2 \theta$$ Recalling that $i^2 = -1$, the expression simplifies to: $$(2 - 6 \sin^2 \theta) + i(7 \sin \theta)$$ ### Solving for the Condition We now express $Z$ in the standard form $Z = A + iB$: $$Z = \frac{2 - 6 \sin^2 \theta}{1 + 4 \sin^2 \theta} + i \left( \frac{7 \sin \theta}{1 + 4 \sin^2 \theta} \right)$$ For $Z$ to be **purely imaginary**, the real part $A$ must be zero: $$\frac{2 - 6 \sin^2 \theta}{1 + 4 \sin^2 \theta} = 0$$ This implies: $$2 - 6 \sin^2 \theta = 0 \Rightarrow \sin^2 \theta = \frac{1}{3}$$ Thus, the condition for the complex number to be purely imaginary is: $$\sin \theta = \pm \frac{1}{\sqrt{3}}$$

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Animated Solution for Mathematics - Complex Numbers: A value of $θ$ for which $\frac{2 + 3 i s i n θ}{1 - 2 i s i n θ}$ is purely imaginary, is

Q: Why do we only take the positive root for $\sin \theta$?

Great observation. Mathematically, $\sin^2 \theta = 1/3$ implies $\sin \theta = \pm 1/\sqrt{3}$. However, in JEE multiple-choice questions, we look for the value that matches the provided options. Since $1/\sqrt{3}$ is listed, that is our target.

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Visualized Solution

The Complex Plane

A complex number $Z = x + i y$ can be plotted on the Argand plane.
The x-axis represents the Real part, and the y-axis represents the Imaginary part.

Condition for Purely Imaginary

For $Z$ to be purely imaginary, it must lie exactly on the imaginary axis.
This means its horizontal component must be zero: $Re (Z) = 0$ .

Defining the Expression

Let the given complex number be $Z$ .
$Z = \frac{2 + 3 i s i n θ}{1 - 2 i s i n θ}$

Rationalizing the Denominator

To separate the real and imaginary parts, we must rationalize the denominator.
Multiply the numerator and denominator by the complex conjugate of the denominator.
Conjugate of $1 - 2 i sin θ$ is $1 + 2 i sin θ$ .

Setting up the Multiplication

$Z = \frac{2 + 3 i s i n θ}{1 - 2 i s i n θ} \times \frac{1 + 2 i s i n θ}{1 + 2 i s i n θ}$

Expanding the Numerator

Let's expand the numerator step-by-step:
$(2 + 3 i sin θ) (1 + 2 i sin θ)$
$= 2 (1) + 2 (2 i sin θ) + 3 i sin θ (1) + 3 i sin θ (2 i sin θ)$

Simplifying the Numerator

$= 2 + 4 i sin θ + 3 i sin θ + 6 i^{2} sin^{2} θ$
Since $i^{2} = - 1$ , the last term becomes $- 6 sin^{2} θ$ .

Grouping Real and Imaginary Parts

Grouping the real terms and imaginary terms in the numerator:
$Numerator = (2 - 6 sin^{2} θ) + i (7 sin θ)$

Expanding the Denominator

Now for the denominator: $(1 - 2 i sin θ) (1 + 2 i sin θ)$
Using $(a - b) (a + b) = a^{2} - b^{2}$ :
$= (1)^{2} - (2 i sin θ)^{2} = 1 - 4 i^{2} sin^{2} θ$
$= 1 + 4 sin^{2} θ$

The Standard Form of $Z$

Dividing both parts of the numerator by the denominator:
$Z = \frac{2 - 6 s i n ^{2} θ}{1 + 4 s i n ^{2} θ} + i (\frac{7 s i n θ}{1 + 4 s i n ^{2} θ})$

Applying the Purely Imaginary Condition

We know that for $Z$ to be purely imaginary, $Re (Z) = 0$ .
$\frac{2 - 6 s i n ^{2} θ}{1 + 4 s i n ^{2} θ} = 0$

Solving the Equation

Multiplying both sides by the denominator:
$2 - 6 sin^{2} θ = 0$
$6 sin^{2} θ = 2$

Finding $sin θ$

$sin^{2} θ = \frac{2}{6} = \frac{1}{3}$
Taking the square root:
$sin θ = \frac{1}{3}$ (Considering the principal value from options)

Final Value of $θ$

$θ = sin^{- 1} (\frac{1}{3})$
Key Takeaway: For purely imaginary numbers, always convert to standard form and set the real part to zero.

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The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Understanding the Purely Imaginary Condition

In the Argand plane, a complex number

Z = x + i y

represents a point where

x

is the real part and

y

is the imaginary part.

When we define a number as purely imaginary, we are stating that it has no horizontal component. Geometrically, this means the real part must vanish, so we set

x = 0

Untangling the Expression

We are given the complex expression:

Z = \frac{2 + 3 i sin θ}{1 - 2 i sin θ}

To simplify this, we employ the technique of rationalization. We multiply both the numerator and the denominator by the complex conjugate of the denominator, which is

1 + 2 i sin θ

The denominator becomes:

(1 - 2 i sin θ) (1 + 2 i sin θ) = 1^{2} - (2 i sin θ)^{2} = 1 + 4 sin^{2} θ

Expanding the Numerator

Next, we expand the numerator:

(2 + 3 i sin θ) (1 + 2 i sin θ) = 2 + 4 i sin θ + 3 i sin θ + 6 i^{2} sin^{2} θ

Recalling that

i^{2} = - 1

, the expression simplifies to:

(2 - 6 sin^{2} θ) + i (7 sin θ)

Solving for the Condition

We now express

Z

in the standard form

Z = A + i B

Z = \frac{2 - 6 sin ^{2} θ}{1 + 4 sin ^{2} θ} + i (\frac{7 sin θ}{1 + 4 sin ^{2} θ})

For

Z

to be purely imaginary, the real part

A

must be zero:

\frac{2 - 6 sin ^{2} θ}{1 + 4 sin ^{2} θ} = 0

This implies:

2 - 6 sin^{2} θ = 0 \Rightarrow sin^{2} θ = \frac{1}{3}

Thus, the condition for the complex number to be purely imaginary is:

sin θ = \pm \frac{1}{3}

Common Traps & FAQs

Wait, why can't I just set the numerator of the original expression to zero? Isn't that the real part?

That is a classic trap! The original expression is a ratio. The real part of a quotient is not simply the real part of the numerator. You must rationalize the denominator first to isolate the true real part of the entire complex number.

What if the denominator becomes zero? Do we need to check for that?

Excellent mathematical hygiene! Yes, always check the domain. Here, the denominator is

1 + 4 sin^{2} θ

. Since

sin^{2} θ

is always non-negative,

1 + 4 sin^{2} θ \geq 1

. It can never be zero, so our solution is safe.

Why do we only take the positive root for

sin θ

Great observation. Mathematically,

sin^{2} θ = 1/3

implies

sin θ = \pm 1/ 3

. However, in JEE multiple-choice questions, we look for the value that matches the provided options. Since

1/ 3

is listed, that is our target.

Animated Solution for Mathematics - Complex Numbers: A value of $θ$ for which $\frac{2 + 3 i s i n θ}{1 - 2 i s i n θ}$ is purely imaginary, is

Select Answer:

Visualized Solution

The Complex Plane

Condition for Purely Imaginary

Defining the Expression

Rationalizing the Denominator

Setting up the Multiplication

Expanding the Numerator

Simplifying the Numerator

Grouping Real and Imaginary Parts

Expanding the Denominator

The Standard Form of $Z$

Applying the Purely Imaginary Condition

Solving the Equation

Finding $sin θ$

Final Value of $θ$

The Sigma Insight: Algebraic Operations on Complex Numbers

Understanding the Purely Imaginary Condition

Untangling the Expression

Expanding the Numerator

Solving for the Condition

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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; } A value of θ for which 1−2isinθ2+3isinθ​ is purely imaginary, is

Select Answer:

Visualized Solution

The Complex Plane

Condition for Purely Imaginary

Defining the Expression

Rationalizing the Denominator

Setting up the Multiplication

Expanding the Numerator

Simplifying the Numerator

Grouping Real and Imaginary Parts

Expanding the Denominator

The Standard Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z

Applying the Purely Imaginary Condition

Solving the Equation

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

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

The Sigma Insight: Algebraic Operations on Complex Numbers

Understanding the Purely Imaginary Condition

Untangling the Expression

Expanding the Numerator

Solving for the Condition

Similar Questions

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Animated Solution for Mathematics - Complex Numbers: A value of $θ$ for which $\frac{2 + 3 i s i n θ}{1 - 2 i s i n θ}$ is purely imaginary, is

The Standard Form of $Z$

Finding $sin θ$

Final Value of $θ$

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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

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