$\sin x+i\cos 2x$ and $\cos x-i\sin 2x$ are conjugate to each other for: (A) $x=n\pi$ (B) $x=\left(n+\dfrac{1}{2}\right)\dfrac{\pi}{2}$ (C) $x=0$ (D) No value of $x$

sin x+icos 2x and cos x-isin 2x are conjugate to each other for: (A) x=nπ (B) x=(n+(1)/(2))(π)/(2) (C) x=0 (D)

NCERT Exemplar · Q23

\sin x+i\cos 2x

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Concept understanding — Complex Number Arithmetic

Complex Number Arithmetic: A First Look

Imagine you're trying to solve $x^2 + 1 = 0$ . You know that no real number squared gives $-1$ . The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is $-1$ ? That's exactly what mathematicians did — and that invention is the imaginary unit $i$ , defined by:

i^2 = -1

A complex number is any number of the form $a + bi$ , where $a$ and $b$ are real numbers. Here $a$ is called the real part, and $b$ is called the imaginary part. For example, $3 + 4i$ has real part $3$ and imaginary part $4$ .

Note

The name "imaginary" is unfortunate — these numbers are just as real (in the mathematical sense) as the numbers you already know. They're simply a different kind of number.

Why Bother?

Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.

Arithmetic Operations

The rules are straightforward: treat $i$ like a variable, but remember that $i^2 = -1$ .

Addition and Subtraction

Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) - (c + di) = (a - c) + (b - d)i

Example: $(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$

Multiplication

Multiply like binomials, then replace $i^2$ with $-1$ .

(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

Example: $(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i)$

$= 8 - 10i + 12i - 15i^2$

$= 8 + 2i - 15(-1)$

$= 8 + 2i + 15 = 23 + 2i$

Watch out

The most common mistake: forgetting that $i^2 = -1$ , not $1$ . Always check your final step.

Division

Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of $a + bi$ is $a - bi$ . When you multiply a complex number by its conjugate, you get a real number:

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2

So to divide:

\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}

Example: $\frac{2 + 3i}{4 - 5i} = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{8 + 10i + 12i + 15i^2}{16 + 25} = \frac{8 + 22i - 15}{41} = \frac{-7 + 22i}{41} = -\frac{7}{41} + \frac{22}{41}i$

Tip

To divide quickly: multiply top and bottom by the conjugate of the denominator, then simplify. The denominator always becomes $c^2 + d^2$ , a positive real number.

The Big Picture …

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Two complex numbers are conjugates when their real parts are equal and their imaginary parts are opposite. Setting $\sin x = \cos x$ and $\cos 2x = \sin 2x$ leads to a contradiction, so no $x$ satisfies both — the answer is (D).

We have two complex numbers:

z_1 = \sin x + i\cos 2x \quad \text{and} \quad z_2 = \cos x - i\sin 2x.

For them to be conjugates of each other, we need $z_1 = \overline{z_2}$ . That means the real part of $z_1$ must equal the real part of $z_2$ , and the imaginary part of $z_1$ must be the negative of the imaginary part of $z_2$ .

Let’s write $z_2$ in standard form: its real part is $\cos x$ , its imaginary part is $-\sin 2x$ . So the conjugate of $z_2$ is:

\overline{z_2} = \cos x + i\sin 2x.

The condition $z_1 = \overline{z_2}$ gives us two equations:

Real parts equal:

\sin x = \cos x.

This happens when $\tan x = 1$ , i.e. $x = n\pi + \frac{\pi}{4}$ for integer $n$ .

Imaginary parts equal:

\cos 2x = \sin 2x.

This happens when $\tan 2x = 1$ , i.e. $2x = m\pi + \frac{\pi}{4}$ , so $x = \frac{m\pi}{2} + \frac{\pi}{8}$ .

Now we need a value of $x$ that satisfies both conditions simultaneously. Let’s check if that’s possible.

From the first condition: $x = n\pi + \frac{\pi}{4}$ .

From the second: $x = \frac{m\pi}{2} + \frac{\pi}{8}$ .

Set them equal:

n\pi + \frac{\pi}{4} = \frac{m\pi}{2} + \frac{\pi}{8}.

Multiply through by $8$ to clear denominators:

8n\pi + 2\pi = 4m\pi + \pi.

Divide by $\pi$ :

$8n + 2 = 4m + 1 \quad \Rightarrow \quad 8n + 1 = 4m.$ …

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Q.sin⁡x+icos⁡2x\sin x+i\cos 2xsinx+icos2x and cos⁡x−isin⁡2x\cos x-i\sin 2xcosx−isin2x are conjugate to each other for: (A) x=nπx=n\pix=nπ (B) x=(n+12)π2x=\left(n+\dfrac{1}{2}\right)\dfrac{\pi}{2}x=(n+21​)2π​ (C) x=0x=0x=0 (D) No value of xxx

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Q. $\sin x+i\cos 2x$ and $\cos x-i\sin 2x$ are conjugate to each other for:
(A) $x=n\pi$
(B) $x=\left(n+\dfrac{1}{2}\right)\dfrac{\pi}{2}$
(C) $x=0$
(D) No value of $x$

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