The real value of $\alpha$ for which the expression $\dfrac{1-i\sin\alpha}{1+2i\sin\alpha}$ is purely real is: (A) $(n+1)\dfrac{\pi}{2}$ (B) $(2n+1)\dfrac{\pi}{2}$ (C) $n\pi$ (D) None of these, where $n\in\mathbf{N}$

The real value of α for which the expression (1-isinα)/(1+2isinα) is purely real is: (A) (n+1)(π)/(2) (B) (2n+

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Concept understanding — Complex Number Transform

The Intuition: Why "Transform" a Number?

You already know that a real number lives on a line — the number line. Adding $+2$ slides you right; multiplying by $-1$ flips you to the opposite side. But what if you want to rotate something? A real number can't do that on its own. Multiplying by $-1$ is a 180° rotation, but what about a 90° rotation? That's where the complex number transform comes in.

Think of a complex number $z = a + bi$ as a point (or an arrow) on a 2D plane. The real part $a$ is the horizontal coordinate, the imaginary part $b$ is the vertical coordinate. Now, when you multiply two complex numbers, something beautiful happens: the lengths multiply, and the angles add.

Note

This is the core insight: multiplication of complex numbers is a rotation + scaling operation, not just a scaling like real numbers.

So a "complex number transform" is simply the act of applying a complex number (as an operator) to another complex number (as a point) — usually by multiplication — to achieve a geometric transformation: rotation, scaling, or both.

The Precise Statement

Let $z = x + yi$ be any complex number (the "point" you want to transform).

Let $w = r(\cos\theta + i\sin\theta)$ be a fixed complex number (the "transformer").

Then the complex number transform of $z$ by $w$ is:

w \cdot z = r(\cos\theta + i\sin\theta) \cdot (x + yi)

When you multiply this out (using $i^2 = -1$ ), the result is a new complex number $z'$ whose geometric meaning is:

Scale the distance of $z$ from the origin by a factor of $r$
Rotate the point $z$ around the origin by an angle $\theta$ counterclockwise

\text{If } w = re^{i\theta}, \text{ then } w \cdot z \text{ rotates } z \text{ by } \theta \text{ and scales it by } r.

This is often written using Euler's formula: $e^{i\theta} = \cos\theta + i\sin\theta$ , so $w = re^{i\theta}$ .

A Concrete Example

Take the point $z = 1 + 0i$ (the number 1 on the real axis).

Let the transformer be $w = i$ (which has $r=1$ , $\theta = 90^\circ$ ).

i \cdot 1 = i

The point $(1,0)$ moved to $(0,1)$ — a 90° rotation counterclockwise. No scaling because $|i| = 1$ .

Now take $z = 2 + 0i$ and $w = 2i$ (which has $r=2$ , $\theta = 90^\circ$ ):

2i \cdot 2 = 4i

The point $(2,0)$ moved to $(0,4)$ — rotated 90° and scaled by factor 2.

Watch out

A common mistake: thinking that multiplying by $i$ always gives a 90° rotation. It does — but only if you multiply the entire complex number. Multiplying just the real part by $i$ is not the same as multiplying the whole number.

Why This Matters

This transform is the foundation of: …

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A complex number is purely real if its imaginary part is zero, or equivalently, if it equals its conjugate. Setting the given expression equal to its conjugate leads to $\sin\alpha = 0$ , which means $\alpha = n\pi$ .

Concept and Intuition

A complex number $z$ is said to be purely real if its imaginary part is zero. That is, if $z = x + iy$ , then $z$ is purely real if $y=0$ .

There are two main ways to approach this problem:

Rationalize the denominator: Convert the complex fraction into the standard form $x+iy$ by multiplying the numerator and denominator by the conjugate of the denominator. Then, set the imaginary part $y$ to zero.
Use the property $z = \bar{z}$ : A complex number $z$ is purely real if and only if $z$ is equal to its complex conjugate $\bar{z}$ . This method is often more elegant and less prone to algebraic errors when dealing with complex fractions.

We will use the second method, $z = \bar{z}$ , as it directly leverages a fundamental property of purely real numbers and simplifies the algebra.

z

Step-by-step Derivation

Define the complex expression and its conjugate: Let the given complex expression be $z$ .

z = \dfrac{1-i\sin\alpha}{1+2i\sin\alpha}

The complex conjugate of $z$, denoted $\bar{z}$, is found by changing the sign of every imaginary part. For a fraction $\frac{A}{B}$, its conjugate is $\frac{\bar{A}}{\bar{B}}$.

\bar{z} = \overline{\left(\dfrac{1-i\sin\alpha}{1+2i\sin\alpha}\right)} = \dfrac{\overline{1-i\sin\alpha}}{\overline{1+2i\sin\alpha}} = \dfrac{1+i\sin\alpha}{1-2i\sin\alpha}

Apply the condition for a purely real number: For $z$ to be purely real, we must have $z = \bar{z}$ .

\dfrac{1-i\sin\alpha}{1+2i\sin\alpha} = \dfrac{1+i\sin\alpha}{1-2i\sin\alpha}

Cross-multiply and expand: Multiply both sides by $(1+2i\sin\alpha)(1-2i\sin\alpha)$ to clear the denominators.

(1-i\sin\alpha)(1-2i\sin\alpha) = (1+i\sin\alpha)(1+2i\sin\alpha)

Now, expand both sides using the distributive property (or FOIL method):
Left Hand Side (LHS):

1(1) + 1(-2i\sin\alpha) + (-i\sin\alpha)(1) + (-i\sin\alpha)(-2i\sin\alpha)

= 1 - 2i\sin\alpha - i\sin\alpha + 2i^2\sin^2\alpha

Since $i^2 = -1$:

= 1 - 3i\sin\alpha - 2\sin^2\alpha

Right Hand Side (RHS):

1(1) + 1(2i\sin\alpha) + (i\sin\alpha)(1) + (i\sin\alpha)(2i\sin\alpha)

= 1 + 2i\sin\alpha + i\sin\alpha + 2i^2\sin^2\alpha

Since $i^2 = -1$:

= 1 + 3i\sin\alpha - 2\sin^2\alpha

Equate LHS and RHS and solve for $\sin\alpha$ : Set the expanded LHS equal to the expanded RHS:

1 - 3i\sin\alpha - 2\sin^2\alpha = 1 + 3i\sin\alpha - 2\sin^2\alpha

Notice that the terms $1$ and $-2\sin^2\alpha$ appear on both sides. We can subtract them from both sides:
$$-3i\sin\alpha = 3i\sin\alpha$$ …

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Concept and Intuition

Step-by-step Derivation

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