NCERT Exemplar Solutions Class 11 Maths Complex Numbers and Quadratic Equations — Objective Type Q35–50

NCERT Exemplar Problems, Class 11 Mathematics — Complex Numbers and Quadratic Equations, Objective Type questions 35 to 50. All 16 questions solved.

Sixteen multiple-choice questions. Questions 36, 40 and 48 ask for the values that make a quotient real or of modulus 1; they all reduce to “multiply by the conjugate and set the imaginary part to zero”. Questions 38, 41, 44, 46 and 50 use $|z|^2 = z\bar z$ and that modulus multiplies and divides; 45, 47 and 49 are about the geometry of $|z – a| = |z – b|$ and of arguments; and 35, 37, 39, 42, 43 are quick standard-form or quadrant checks.

One printed answer is wrong: Question 37. Question 41 also has two true options. Both are explained where they occur. The argument-based parts (Questions 42 and 49, and 47’s “same argument”) are no longer in the current NCERT Class 11 chapter, but are solved here.

Key insight. Convert every condition into a statement about the real part and the imaginary part, or about a distance. “Real” means imaginary part $0$; “purely imaginary” means real part $0$; “$|z – a| = |z – b|$” means a perpendicular bisector. After that the options can be tested one by one with a single sample number.

Question 35

$\sin x + i\cos 2x$ and $\cos x – i\sin 2x$ are conjugate to each other for:

Solution. The conjugate of $\sin x + i\cos 2x$ is $\sin x – i\cos 2x$. Matching it with $\cos x – i\sin 2x$ needs both $\sin x = \cos x$ and $\cos 2x = \sin 2x$.

The first gives $\tan x = 1$, so $x = \frac\pi4 + k\pi$. For these $x$, $2x = \frac\pi2 + 2k\pi$, so $\cos 2x = 0$ but $\sin 2x = 1$, and the second equation fails. So the two conditions are never true together.

Question 36

The real value of $\alpha$ for which the expression $\dfrac{1 – i\sin\alpha}{1 + 2i\sin\alpha}$ is purely real is:

Solution. Multiply top and bottom by the conjugate $1 – 2i\sin\alpha$:

$$\frac{(1 – i\sin\alpha)(1 – 2i\sin\alpha)}{1 + 4\sin^2\alpha} = \frac{(1 – 2\sin^2\alpha) – 3i\sin\alpha}{1 + 4\sin^2\alpha}.$$

The imaginary part is $\dfrac{-3\sin\alpha}{1+4\sin^2\alpha}$, which is zero exactly when $\sin\alpha = 0$, that is $\alpha = n\pi$.

Question 37

If $z = x + iy$ lies in the third quadrant, then $\dfrac{\bar z}{z}$ also lies in the third quadrant if:

Solution. Multiply top and bottom by $\bar z$: $\dfrac{\bar z}{z} = \dfrac{\bar z^2}{|z|^2} = \dfrac{(x^2 – y^2) – 2ixy}{x^2 + y^2}$. We need both parts negative:

• imaginary part: $-2xy < 0 \iff xy > 0$, which is already true because $z$ is in the third quadrant ($x < 0$, $y < 0$);
• real part: $x^2 – y^2 < 0 \iff |x| < |y|$. For negative numbers, a smaller absolute value means a larger number, so $y < x$.

So the condition is $y < x < 0$. Test with numbers: $z = -1 – 2i$ gives $\dfrac{\bar z}{z} = \dfrac{(-1+2i)^2}{5} = \dfrac{-3 – 4i}{5}$, third quadrant. $z = -2 – i$ (which has $x < y < 0$) gives $\dfrac{3 – 4i}{5}$, fourth quadrant.

The printed key is wrong here. It prints (B) $x < y < 0$, which is exactly the case in which $\bar z/z$ falls into the fourth quadrant.

Question 38

The value of $(z + 3)(\bar z + 3)$ is equivalent to:

Solution. The conjugate of $z + 3$ is $\bar z + 3$ (the $3$ is real), and $w\bar w = |w|^2$ for any $w$. So the product is $|z + 3|^2$.

Question 39

If $\left(\dfrac{1+i}{1-i}\right)^x = 1$, then:

Solution. $\dfrac{1+i}{1-i} = i$ (Question 3 of the first page), so the equation is $i^x = 1$. Powers of $i$ cycle $i, -1, -i, 1$ with period $4$, and $i^x = 1$ exactly when $x$ is a multiple of $4$. Option (C) $x = 2n$ includes $x = 2$, where $i^2 = -1$, so it is too wide.

Question 40

A real value of $x$ satisfies the equation $\dfrac{3 – 4ix}{3 + 4ix} = \alpha – i\beta$ ($\alpha, \beta \in \mathbf R$) if $\alpha^2 + \beta^2 =$

Solution. $\alpha^2 + \beta^2 = |\alpha – i\beta|^2$. For real $x$, the numerator $3 – 4ix$ is the conjugate of the denominator $3 + 4ix$, so they have the same modulus and the quotient has modulus $1$. Therefore $\alpha^2 + \beta^2 = 1$. (A sum of squares cannot be $-1$ or $-2$, which also rules out (B) and (D).)

Question 41

Which of the following is correct for any two complex numbers $z_1$ and $z_2$?

Solution. (A) is the multiplicative property of the modulus and always holds. (B) is false: arguments add, they are not multiplied. (C) is false: in general only $|z_1 + z_2| \le |z_1| + |z_2|$ (equal only for the same argument).

(D) is also a true statement. The triangle inequality gives $|z_1 + z_2| \ge \big||z_1| – |z_2|\big| \ge |z_1| – |z_2|$. So the question has two correct options. The key prints (A), the fundamental identity and the intended answer, so (A) is marked.

Question 42

The point represented by the complex number $2 – i$ is rotated about the origin through an angle $\dfrac\pi2$ in the clockwise direction; the new position of the point is:

Solution. Rotating by $\frac\pi2$ clockwise multiplies by $-i$ (the opposite of the anticlockwise rotation, which multiplies by $i$):

$$(2 – i)(-i) = -2i + i^2 = -1 – 2i.$$

Check: $2 – i$ is in the fourth quadrant; a clockwise right-angle turn takes it into the third quadrant, where $-1 – 2i$ lies.

Question 43

Let $x, y \in \mathbf R$. Then $x + iy$ is a non-real complex number if:

Solution. A complex number is real exactly when its imaginary part $y$ is zero, so it is non-real exactly when $y \ne 0$. The real part plays no role: $x = 0$ with $y \ne 0$ is a purely imaginary (non-real) number.

Question 44

If $a + ib = c + id$, then:

Solution. Presumably $a, b, c, d$ are real. Equal complex numbers have equal real and imaginary parts: $a = c$ and $b = d$. Hence $a^2 + b^2 = c^2 + d^2$. The other options are not implied: for $a = c = 1$, $b = d = 0$, options (A), (B) and (C) are false.

Question 45

The complex number $z$ which satisfies the condition $\left|\dfrac{i + z}{i – z}\right| = 1$ lies on:

Solution. The condition says $|z + i| = |z – i|$ (since $|i – z| = |z – i|$): $z$ is equally far from $(0, -1)$ and $(0, 1)$, so it is on the perpendicular bisector of that segment, the $x$-axis. Algebraically, $x^2 + (y+1)^2 = x^2 + (y-1)^2$ gives $4y = 0$.

Question 46

If $z$ is a complex number, then:

Solution. The modulus of a product is the product of moduli, so $|z^2| = |z\cdot z| = |z|\cdot|z| = |z|^2$. It is an equality always; (D) is true only because equality is allowed, but (B) is the exact statement.

Question 47

$|z_1 + z_2| = |z_1| + |z_2|$ is possible if:

Solution. Equality holds in the triangle inequality exactly when the two points lie on the same ray from the origin, that is, when $\arg z_1 = \arg z_2$ (or one of them is $0$). Then $|z_1 + z_2| = (r_1 + r_2)$ because the lengths simply add.

The others do not guarantee it: with $z_1 = i$, $z_2 = \bar z_1 = -i$ the sum is $0$ while $|z_1| + |z_2| = 2$; and $|z_1| = |z_2|$ alone is satisfied by $1$ and $-1$ (sum $0$ against $2$). Options (A) and (B) can hold only in special cases (for instance $z_1 = 1$), which is not a condition in general, so (C) is the answer.

Question 48

The real value of $\theta$ for which the expression $\dfrac{1 + i\cos\theta}{1 – 2i\cos\theta}$ is a real number is:

Solution. Multiply top and bottom by $1 + 2i\cos\theta$:

$$\frac{(1 + i\cos\theta)(1 + 2i\cos\theta)}{1 + 4\cos^2\theta} = \frac{(1 – 2\cos^2\theta) + 3i\cos\theta}{1 + 4\cos^2\theta}.$$

The imaginary part vanishes when $\cos\theta = 0$, so $\theta = (2n+1)\dfrac\pi2$, and the set $\left\{\ldots, -\frac\pi2, \frac\pi2, \frac{3\pi}2, \ldots\right\}$ is the same as $2n\pi \pm \frac\pi2$.

Question 49

The value of $\arg(x)$ when $x < 0$ is:

Solution. A negative real number lies on the negative real axis, at an angle of $\pi$ from the positive real axis. The principal argument (in $(-\pi, \pi]$) is $\pi$.

Question 50

If $f(z) = \dfrac{7 – z}{1 – z^2}$, where $z = 1 + 2i$, then $|f(z)|$ is:

Solution. For $z = 1 + 2i$: $z^2 = 1 – 4 + 4i = -3 + 4i$, so $1 – z^2 = 4 – 4i$ and $7 – z = 6 – 2i$. Then

$$|f(z)| = \frac{|6 – 2i|}{|4 – 4i|} = \frac{\sqrt{40}}{\sqrt{32}} = \sqrt{\frac54} = \frac{\sqrt5}{2}.$$

Since $|z| = \sqrt5$, this is $\dfrac{|z|}{2}$.

Common mistakes

• Question 35: solving only $\sin x = \cos x$, finding $x = \frac\pi4$, and ticking an option. The imaginary parts must also match, and for those $x$ they do not.
• Question 36 and 48: setting the real part to zero, or forgetting to multiply by the conjugate and trying to read the imaginary part off the unsimplified fraction. A quotient is only readable after the denominator is made real.
• Question 37: comparing $|x|$ and $|y|$ and then writing the inequality for the signed numbers the wrong way round. For negative numbers, $|x| < |y|$ means $y < x$. Substituting $-1 – 2i$ settles it in seconds.
• Question 39: picking $x = 2n$ because $i^2 = -1$ “looks like $1$ in size”. The sign matters: $i^2 = -1 \ne 1$.
• Question 41: thinking $\arg(z_1z_2) = \arg z_1\cdot\arg z_2$. Arguments add on multiplication, exactly as exponents do, because $z = r e^{i\theta}$.
• Question 42: multiplying by $i$ instead of $-i$ for a clockwise turn, giving $1 + 2i$ (option (A)), which is the anticlockwise rotation.

Practise next

• The other pages of this unit: Short and Long Answer, Q1–24 and Fill in the Blanks, True or False and Matching, Q25–34.
• NCERT Class 11 Maths Exercise 4.1 — Complex Numbers: standard form, conjugates and modulus.
• NCERT Class 11 Maths Miscellaneous Exercise — Complex Numbers: modulus and conjugate identities used in Questions 38, 40 and 46.