MathsCBSE Class 11 · Chapter 4

Complex Numbers and Quadratic Equations Class 11: Notes and Solved Questions

Updated 14 October 2026 · Fact-checked

Complex Numbers and Quadratic Equations is Chapter 4 of CBSE Class 11 Maths. It explains why we need $i = \sqrt{-1}$, complex numbers in the form a + ib, their addition, subtraction, multiplication and division, powers of i, square roots of negative numbers, the conjugate and modulus, the multiplicative inverse, and the Argand plane.

What Complex Numbers and Quadratic Equations is about

The equation $x^{2} + 1 = 0$ has no real solution, because the square of a real number is never negative. To solve such equations we bring in a new number, $i$ , with $i^{2} = - 1$ . Numbers of the form $z = a + ib$ , with a and b real, are complex numbers: a is the real part and b the imaginary part.

Complex numbers follow the familiar rules of algebra. You add and subtract them part by part, multiply them like binomials using $i^{2} = - 1$ , and divide by multiplying top and bottom by the conjugate $\overset{z}{ˉ} = a - ib$ . Powers of i repeat in a cycle of four: $i, - 1, - i, 1$ . The modulus $∣ z ∣ = a^{2} + b^{2}$ measures the size of z, and $z \overset{z}{ˉ} = ∣ z ∣^{2}$ is the key to division and to the multiplicative inverse $z^{- 1} = \frac{z ˉ}{∣ z ∣ ^{2}}$ .

Finally, the Argand plane pictures $a + ib$ as the point (a, b): the modulus becomes a distance and the conjugate becomes a reflection in the real axis. For 2026-27, polar form and solving quadratic equations with complex roots are taught but assessed only formatively; the NCERT sections on quadratic equations and square roots of a complex number have been removed.

Weightage: This chapter is part of Unit II, Algebra, which the syllabus gives 25 of the 80 theory marks in the school annual exam (CBSE adds that there is no fixed chapter-wise weightage).Source: CBSE Mathematics (041) syllabus 2026-27, Class XI course structure, Unit II Algebra

Key concepts

The number i and complex numbers

NCERT section 4.1 and 4.2

Since no real number squares to −1, we introduce the symbol $i$ (iota) with $i^{2} = - 1$ . So $i = - 1$ , and $x^{2} + 1 = 0$ now has the solutions $\pm i$ .

A complex number is any number $z = a + ib$ with $a, b \in R$ :

$a = Re z$ is the real part; $b = Im z$ is the imaginary part (b itself is real; it is not 'bi').
If $b = 0$ , z is real; if $a = 0$ and $b \neq = 0$ , z is purely imaginary.
Every real number is a complex number, so $R \subset C$ , where C is the set of complex numbers.

Equality: $a + ib = c + i d$ exactly when $a = c$ and $b = d$ . One complex equation gives two real equations.

There is no 'greater than' between complex numbers: $2 + 3 i > 1 + i$ has no meaning.

Worked example

Find real x and y if $(2 x - 1) + i (y + 3) = 5 - 2 i$ . [2 marks]

Try it first, then show the solutionSolution

Equate real parts: $2 x - 1 = 5 \Rightarrow x = 3$
Equate imaginary parts: $y + 3 = - 2 \Rightarrow y = - 5$

Answer: x = 3, y = −5

Addition, subtraction and multiplication

NCERT section 4.3.1 to 4.3.3 and 4.3.7

For $z_{1} = a + ib$ and $z_{2} = c + i d$ :

Sum: $z_{1} + z_{2} = (a + c) + i (b + d)$
Difference: $z_{1} - z_{2} = (a - c) + i (b - d)$
Product: $z_{1} z_{2} = (a c - b d) + i (a d + b c)$ (expand like a binomial, then replace $i^{2}$ by −1)

Properties (same as real numbers): addition and multiplication are closed, commutative and associative; 0 is the additive identity and 1 the multiplicative identity; $- z = - a - ib$ is the additive inverse; multiplication distributes over addition.

Because the laws are the same, real identities hold for complex numbers too: $(z_{1} + z_{2})^{2} = z_{1}^{2} + 2 z_{1} z_{2} + z_{2}^{2}$ , $(z_{1} - z_{2})^{3} = z_{1}^{3} - 3 z_{1}^{2} z_{2} + 3 z_{1} z_{2}^{2} - z_{2}^{3}$ , $z_{1}^{2} - z_{2}^{2} = (z_{1} + z_{2}) (z_{1} - z_{2})$ .

Worked example

Express $(2 + i)^{3}$ in the form a + ib. [2 marks]

Try it first, then show the solutionSolution

$(2 + i)^{3} = 8 + 3 (4) (i) + 3 (2) (i^{2}) + i^{3} = 8 + 12 i - 6 - i$
$= 2 + 11 i$

Answer: $2 + 11 i$

Powers of i

NCERT section 4.3.5

$i^{1} = i, i^{2} = - 1, i^{3} = - i, i^{4} = 1$

Then the pattern repeats every four powers. To find $i^{n}$ , divide n by 4 and use the remainder r: $i^{n} = i^{r}$ (with $i^{0} = 1$ ).

For negative powers, use $i^{- 1} = \frac{1}{i} = - i$ , or add a multiple of 4 to make the power non-negative: $i^{- 39} = i^{- 39 + 40} = i$ .

Useful fact: the sum of any four consecutive powers is zero: $i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} = i^{n} (1 + i - 1 - i) = 0$ .

Worked example

Evaluate $i^{57} + i^{- 25}$ and $i^{35}$ . [2 marks]

Try it first, then show the solutionSolution

57 = 4 × 14 + 1, so $i^{57} = i$ ; $- 25 + 28 = 3$ , so $i^{- 25} = i^{3} = - i$ . Sum = 0
35 = 4 × 8 + 3, so $i^{35} = i^{3} = - i$

Answer: 0 and −i

Square roots of negative real numbers

NCERT section 4.3.6

Both $i$ and $- i$ square to −1, so −1 has two square roots, but the symbol $- 1$ means $i$ only. In general, for $a > 0$ :
$- a = a i$
so $- 9 = 3 i$ , and the two square roots of −9 are $\pm 3 i$ .

Caution: the rule $a b = ab$ holds if at least one of a, b is non-negative, but fails when both are negative. For example $- 4 \times - 9 = 2 i \times 3 i = 6 i^{2} = - 6$ , whereas $(- 4) (- 9) = 36 = 6$ .

Safe habit: convert every $- a$ to $a i$ before multiplying.

Worked example

Simplify $- 16 + 3 - 25 - - 4 \times - 9$ . [2 marks]

Try it first, then show the solutionSolution

$- 16 = 4 i$ , $3 - 25 = 15 i$
$- 4 \times - 9 = 2 i \times 3 i = - 6$ , so the expression is $4 i + 15 i + 6 = 6 + 19 i$

Answer: $6 + 19 i$

Conjugate and modulus

NCERT section 4.4

For $z = a + ib$ :

Conjugate $\overset{z}{ˉ} = a - ib$ (change the sign of the imaginary part).
Modulus $∣ z ∣ = a^{2} + b^{2}$ , a non-negative real number.

Key facts:

$z \overset{z}{ˉ} = a^{2} + b^{2} = ∣ z ∣^{2}$ (always real and non-negative)
$z + \overset{z}{ˉ} = 2 Re z$ , $z - \overset{z}{ˉ} = 2 i Im z$ ; $\overset{ˉ}{\overset{z}{ˉ}} = z$
$∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣ ∣ z_{2} ∣$ and $\frac{z _{1}}{z _{2}} = \frac{∣ z _{1} ∣}{∣ z _{2} ∣}$ (for $z_{2} \neq = 0$ )
$\overline{z_{1} \pm z_{2}} = \overset{z_{1}}{ˉ} \pm \overset{z_{2}}{ˉ}$ , $\overline{z_{1} z_{2}} = \overset{z_{1}}{ˉ} \overset{z_{2}}{ˉ}$ , $\overline{(\frac{z _{1}}{z _{2}})} = \frac{z _{1} ˉ}{z _{2} ˉ}$
z is real exactly when $z = \overset{z}{ˉ}$ ; purely imaginary (or 0) exactly when $z = - \overset{z}{ˉ}$ .

Multiplicative inverse: for $z \neq = 0$ , $z^{- 1} = \frac{z ˉ}{∣ z ∣ ^{2}} = \frac{a - ib}{a ^{2} + b ^{2}}$ , because $z \cdot \frac{z ˉ}{∣ z ∣ ^{2}} = 1$ .

Worked example

Find the multiplicative inverse of $3 + 4 i$ and check your answer. [2 marks]

Try it first, then show the solutionSolution

$∣3 + 4 i ∣^{2} = 9 + 16 = 25$ , so $z^{- 1} = \frac{3 - 4 i}{25} = \frac{3}{25} - \frac{4}{25} i$
Check: $(3 + 4 i) (3 - 4 i) = 9 + 16 = 25$ , and $\frac{25}{25} = 1$ ✓

Answer: $\frac{3}{25} - \frac{4}{25} i$

Division of complex numbers

NCERT section 4.3.4 and 4.4

To divide, multiply the top and bottom by the conjugate of the denominator, which makes the denominator a positive real number:
$\frac{a + ib}{c + i d} = \frac{( a + ib ) ( c - i d )}{c ^{2} + d ^{2}}$

Then split into real and imaginary parts. This is the same idea as rationalising a surd denominator.

If you only need the modulus of a quotient, use $\frac{z _{1}}{z _{2}} = \frac{∣ z _{1} ∣}{∣ z _{2} ∣}$ ; there is no need to divide first.

Worked example

Express $\frac{3 + 2 i}{1 - i}$ in the form a + ib. [2 marks]

Try it first, then show the solutionSolution

Multiply top and bottom by $1 + i$ : $\frac{( 3 + 2 i ) ( 1 + i )}{1 + 1} = \frac{3 + 3 i + 2 i + 2 i ^{2}}{2}$
$= \frac{1 + 5 i}{2} = \frac{1}{2} + \frac{5}{2} i$

Answer: $\frac{1}{2} + \frac{5}{2} i$

The Argand plane

NCERT section 4.5

Each complex number $x + i y$ matches exactly one point P(x, y) of the coordinate plane, and each point matches exactly one complex number. A plane used this way is the complex plane or Argand plane.

The x-axis is the real axis (numbers $a + 0 i$ ); the y-axis is the imaginary axis (numbers $0 + ib$ ).
$∣ z ∣ = x^{2} + y^{2}$ is the distance of P from the origin; $∣ z_{1} - z_{2} ∣$ is the distance between the two points.
The conjugate $\overset{z}{ˉ} = x - i y$ is the point (x, −y), the mirror image of P in the real axis.
Multiplying by i turns a point through 90° anticlockwise about the origin: $i (x + i y) = - y + i x$ .

(Polar form $z = r (cos θ + i sin θ)$ is taught but assessed only formatively in 2026-27.)

Worked example

Plot $z = - 2 + 3 i$ and $\overset{z}{ˉ}$ in the Argand plane. State the quadrant of each and the distance of z from the origin. [2 marks]

Try it first, then show the solutionSolution

z is the point (−2, 3), in the second quadrant; $\overset{z}{ˉ} = - 2 - 3 i$ is (−2, −3), in the third quadrant, its mirror image in the real axis
$∣ z ∣ = 4 + 9 = 13$

Answer: Quadrants II and III; distance $13$ .

Maths in real life: Complex Numbers and Quadratic Equations

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Rotation by multiplying by i

Suppose a drone hovers at the point that matches $2 + 3 i$ in an Argand plane drawn on a playground, with the origin at the controller.

The pilot's software multiplies the position by i, then by i again. Where is the drone after each step, and what does each step do geometrically?

Try it first, then show the solutionSolution

$i (2 + 3 i) = 2 i + 3 i^{2} = - 3 + 2 i$ : the point (−3, 2)
$i (- 3 + 2 i) = - 3 i - 2 = - 2 - 3 i$ : the point (−2, −3)
Each multiplication by i turns the position through 90° anticlockwise about the origin, keeping the distance $13$

Answer: (−3, 2), then (−2, −3); two quarter turns make a half turn

Memory hook: Times i = quarter turn; times −1 = half turn.

Modulus as size

Imagine an electrical engineer who describes the opposition of a circuit component to alternating current by a complex number called impedance, here $Z = 3 + 4 i$ ohms.

The size of the opposition is $∣ Z ∣$ . Find it. If a second component with $Z_{2} = 1 - 2 i$ is joined in series, the impedances add. Find the total and its size.

Try it first, then show the solutionSolution

$∣ Z ∣ = 9 + 16 = 5$ ohms
Series total: $(3 + 4 i) + (1 - 2 i) = 4 + 2 i$
$∣4 + 2 i ∣ = 16 + 4 = 20 \approx 4.47$ ohms

Answer: 5 ohms; total $4 + 2 i$ with size about 4.47 ohms

Memory hook: Modulus = distance from 0 = Pythagoras on (a, b).

Why complex numbers are needed

Suppose a student models a design problem and ends up with the equation $x^{2} + 4 = 0$ .

Explain why there is no real answer and write the two complex solutions.

Try it first, then show the solutionSolution

For real x, $x^{2} \geq 0$ , so $x^{2} + 4 \geq 4 > 0$ : no real solution
$x^{2} = - 4 \Rightarrow x = \pm - 4 = \pm 2 i$ ; check: $(2 i)^{2} + 4 = - 4 + 4 = 0$

Answer: $x = 2 i$ or $x = - 2 i$

Memory hook: Negative under the square root? Pull out an i.

Topics in this chapter

Complex Numbers and Quadratic Equations formula sheet

Iota and powers: $i^{2} = - 1, i^{3} = - i, i^{4} = 1, i^{4 k + r} = i^{r}$
Product: $(a + ib) (c + i d) = (a c - b d) + i (a d + b c)$
Quotient: $\frac{a + ib}{c + i d} = \frac{( a c + b d ) + i ( b c - a d )}{c ^{2} + d ^{2}}$
Conjugate and modulus: $\overset{z}{ˉ} = a - ib, ∣ z ∣ = a^{2} + b^{2}, z \overset{z}{ˉ} = ∣ z ∣^{2}$
Multiplicative inverse: $z^{- 1} = \frac{z ˉ}{∣ z ∣ ^{2}} = \frac{a - ib}{a ^{2} + b ^{2}}, z \neq = 0$
Modulus rules: $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣, \frac{z _{1}}{z _{2}} = \frac{∣ z _{1} ∣}{∣ z _{2} ∣}$
Square root of a negative number: $- a = a i (a > 0)$; $a b \neq = ab$ when a and b are both negative.

How the school exam asks Complex Numbers and Quadratic Equations

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Standard form a + ib	Simplify sums, products, powers and quotients into a + ib (Exercise 4.1 style).	Expand, replace $i^{2}$ by −1, collect real and imaginary parts; for quotients multiply by the conjugate.
Powers of i	Evaluate $i^{n}$ , negative powers and sums of powers, e.g. $(i^{22} + (\frac{1}{i})^{27})^{3}$ .	Reduce the power mod 4; turn negative powers into positive ones by adding multiples of 4.
Equality of complex numbers	Find real x and y from an equation such as $(x + i y) (1 + 2 i) = 5 + 5 i$ .	Bring to the form a + ib = c + id and equate real and imaginary parts.
Conjugate, modulus and inverse	Find $\overset{z}{ˉ}$ , $∣ z ∣$ , $z^{- 1}$ ; prove results like $Re (z_{1} z_{2}) = Re z_{1} Re z_{2} - Im z_{1} Im z_{2}$ .	Use $z \overset{z}{ˉ} = ∣ z ∣^{2}$ and the modulus rules; write $z_{1}, z_{2}$ in a + ib form for proofs.
Argand plane	Plot numbers and their conjugates; interpret modulus as distance.	a + ib ↔ (a, b); conjugate = reflection in the real axis.

Tricks to remember Complex Numbers and Quadratic Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

i, −1, −i, 1 on repeat
Powers of i go round a clock of four: divide the power by 4 and use the remainder.
Use it when: Any power of i.
Conjugate kills the i
$(a + ib) (a - ib) = a^{2} + b^{2}$ : a real number. Use it to clear denominators.
Use it when: Division and inverses.
Root first, multiply later
Change $- a$ to $a i$ before any multiplication.
Use it when: Avoiding the $- 4 - 9 = 6$ trap.
Modulus is a distance
$∣ a + ib ∣$ is how far (a, b) is from the origin: Pythagoras.
Use it when: Moduli and the Argand plane.
Bar = flip in the real axis
The conjugate reflects the point in the x-axis.
Use it when: Plotting conjugates.

Complex Numbers and Quadratic Equations: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Power of i · 1 mark · MCQExam level

The value of $i^{- 42}$ is:
(a) 1
(b) −1
(c) i
(d) −i

Try it first, then show the solutionSolution

$i^{- 42} = i^{- 42 + 44} = i^{2} = - 1$ .

Answer: (b) −1

Example 2: The square-root trap · 1 mark · Assertion-reasonExam level

Assertion (A): $- 2 \times - 3 = 6$ .
Reason (R): $a b = ab$ for all real a and b.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) Both A and R false.

Try it first, then show the solutionSolution

$- 2 - 3 = 2 i \cdot 3 i = - 6$ , so A is false. R fails when both a and b are negative, so R is false too.

Answer: (d)

Example 3: Conjugate of a quotient · 3 marks · Short answerHarder exam level

Find the conjugate of $\frac{( 2 + i ) ^{2}}{3 - i}$ . [3 marks]

Try it first, then show the solutionSolution

$(2 + i)^{2} = 4 + 4 i - 1 = 3 + 4 i$
$\frac{3 + 4 i}{3 - i} = \frac{( 3 + 4 i ) ( 3 + i )}{10} = \frac{9 + 3 i + 12 i - 4}{10} = \frac{5 + 15 i}{10} = \frac{1}{2} + \frac{3}{2} i$
Conjugate $= \frac{1}{2} - \frac{3}{2} i$

Answer: $\frac{1}{2} - \frac{3}{2} i$

Example 4: Modulus of a sum of quotients · 2 marks · Very short answerHarder exam level

Find the modulus of $\frac{2 + i}{2 - i} + \frac{2 - i}{2 + i}$ . [2 marks]

Try it first, then show the solutionSolution

$\frac{2 + i}{2 - i} = \frac{( 2 + i ) ^{2}}{5} = \frac{3 + 4 i}{5}$ ; similarly $\frac{2 - i}{2 + i} = \frac{3 - 4 i}{5}$
Sum $= \frac{6}{5}$ , a positive real number, so its modulus is $\frac{6}{5}$

Answer: $\frac{6}{5}$

Example 5: A large power · 3 marks · Short answerHarder exam level

Show that $(1 + i)^{8} = 16$ and hence find $(1 + i)^{10}$ . [3 marks]

Try it first, then show the solutionSolution

$(1 + i)^{2} = 1 + 2 i - 1 = 2 i$
$(1 + i)^{8} = (2 i)^{4} = 16 i^{4} = 16$
$(1 + i)^{10} = 16 \times 2 i = 32 i$

Answer: 16; $32 i$

Example 6: Proof with real and imaginary parts · 3 marks · Short answerHarder exam level

For any complex numbers $z_{1}$ and $z_{2}$ , prove that $Im (z_{1} z_{2}) = Re z_{1} Im z_{2} + Im z_{1} Re z_{2}$ . [3 marks]

Try it first, then show the solutionSolution

Let $z_{1} = a + ib$ and $z_{2} = c + i d$
$z_{1} z_{2} = (a c - b d) + i (a d + b c)$ , so $Im (z_{1} z_{2}) = a d + b c$
$a d + b c = Re z_{1} Im z_{2} + Im z_{1} Re z_{2}$

Answer: Proved.

Example 7: Unit modulus · 3 marks · Short answerHOTS / competency

If $∣ z ∣ = 1$ and $z \neq = - 1$ , show that $\frac{z - 1}{z + 1}$ is purely imaginary or zero. [3 marks]

Try it first, then show the solutionSolution

$∣ z ∣ = 1$ gives $z \overset{z}{ˉ} = 1$ , so $\overset{z}{ˉ} = \frac{1}{z}$
Let $w = \frac{z - 1}{z + 1}$ . Then $\overset{w}{ˉ} = \frac{z ˉ - 1}{z ˉ + 1} = \frac{\frac{1}{z} - 1}{\frac{1}{z} + 1} = \frac{1 - z}{1 + z} = - w$
$\overset{w}{ˉ} = - w$ means Re w = 0, so w is purely imaginary (or zero when z = 1)

Answer: Proved.

Example 8: Sum of powers · 2 marks · Very short answerHOTS / competency

Find the value of $1 + i^{2} + i^{4} + i^{6} + \dots + i^{20}$ . [2 marks]

Try it first, then show the solutionSolution

The terms are $i^{0}, i^{2}, \dots, i^{20}$ : 11 terms that alternate 1, −1, 1, −1, …
Five pairs (1 − 1) give 0, and the last term $i^{20} = 1$ remains

Answer: 1

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Robot on a grid

Suppose a robot's position on a factory floor is recorded as a complex number, with the charging dock at 0. The robot starts at $z = 4 + 3 i$ (metres east and north).

(i) · 1 mark
How far is the robot from the dock?
Answer
Answer: 5 m
$∣4 + 3 i ∣ = 16 + 9 = 5$
(ii) · 1 mark
A command multiplies the position by i. Where does the robot end up?
Answer
Answer: $- 3 + 4 i$
$i (4 + 3 i) = 4 i - 3$
(iii) · 1 mark
A mirror command replaces z by $\overset{z}{ˉ}$ . Where would $4 + 3 i$ go?
Answer
Answer: $4 - 3 i$
Reflection in the real (east-west) axis.
(iv) · 1 mark
How far apart are $4 + 3 i$ and $- 3 + 4 i$ ?
Answer
Answer: $52$ m
$∣ (4 + 3 i) - (- 3 + 4 i) ∣ = ∣7 - i ∣ = 50 = 52$

Checking a classmate's work

Suppose a classmate writes: ' $- 9 \times - 4 = 36 = 6$ , and $\frac{1}{2 + i} = \frac{1}{2} + \frac{1}{i} = \frac{1}{2} - i$ .'

(i) · 1 mark
What is the correct value of $- 9 \times - 4$ ?
Answer
Answer: −6
$3 i \times 2 i = 6 i^{2} = - 6$
(ii) · 1 mark
What is wrong with the second line?
Answer
Answer: $\frac{1}{a + b} \neq = \frac{1}{a} + \frac{1}{b}$
Reciprocals do not split over a sum.
(iii) · 1 mark
Find $\frac{1}{2 + i}$ correctly.
Answer
Answer: $\frac{2}{5} - \frac{1}{5} i$
$\frac{2 - i}{4 + 1}$
(iv) · 1 mark
Check your answer by multiplying by $2 + i$ .
Answer
Answer: Product = 1
$\frac{( 2 - i ) ( 2 + i )}{5} = \frac{5}{5} = 1$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Evaluate $(i^{22} + (\frac{1}{i})^{27})^{3}$ . [2 marks] (2 marks)

Step	Marks
$i^{22} = i^{2} = - 1$ ; $(\frac{1}{i})^{27} = (- i)^{27} = - i^{27} = - i^{3} = i$	1
$(- 1 + i)^{2} = 1 - 2 i + i^{2} = - 2 i$ , so $(- 1 + i)^{3} = - 2 i (- 1 + i) = 2 i - 2 i^{2} = 2 + 2 i$	1

Find real numbers x and y if $(x + i y) (1 + 2 i) = 5 + 5 i$ . [3 marks] (3 marks)

Step	Marks
$x + i y = \frac{5 + 5 i}{1 + 2 i} = \frac{( 5 + 5 i ) ( 1 - 2 i )}{1 + 4}$	1
$= \frac{5 - 10 i + 5 i + 10}{5} = \frac{15 - 5 i}{5} = 3 - i$	1
Equating parts: x = 3, y = −1	1

Let $z_{1} = 3 + i$ and $z_{2} = 1 - 2 i$ . (a) Find $z_{1} z_{2}$ and verify that $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ . (b) Verify that $Re (z_{1} z_{2}) = Re z_{1} Re z_{2} - Im z_{1} Im z_{2}$ . (c) Find $\frac{z _{1} + z _{2} + 1}{z _{1} - z _{2}}$ . [5 marks] (5 marks)

Step	Marks
(a) $z_{1} z_{2} = (3 + i) (1 - 2 i) = 3 - 6 i + i - 2 i^{2} = 5 - 5 i$	1
$∣5 - 5 i ∣ = 50 = 52$ ; $∣ z_{1} ∣∣ z_{2} ∣ = 10 \cdot 5 = 50 = 52$ ✓	1
(b) Re z₁ Re z₂ − Im z₁ Im z₂ = 3(1) − (1)(−2) = 5 = Re(z₁z₂) ✓	1
(c) $z_{1} + z_{2} + 1 = 5 - i$ ; $z_{1} - z_{2} = 2 + 3 i$	1
$\frac{5 - i}{2 + 3 i} = \frac{26}{13} = 2$	1

Where students lose marks in Complex Numbers and Quadratic Equations

The slips examiners see most often, and what to write instead.

Mistake: Taking Im z as 'ib' instead of b
Including i in the imaginary part.
Wrong working
Im(3 − 4i) = −4i
Right working
Im(3 − 4i) = −4
Do this: Both Re z and Im z are real numbers.
Usually costs: 1 mark.
Mistake: Multiplying square roots of two negatives directly
Using $a b = ab$ outside its range.
Wrong working
$- 4 - 9 = 36 = 6$
Right working
$2 i \times 3 i = - 6$
Do this: Convert to i form first.
Usually costs: 1 mark.
Mistake: Forgetting $i^{2} = - 1$ in a product
Expanding like real algebra and stopping.
Do this: After expanding, replace every $i^{2}$ by −1 and every $i^{3}$ by −i.
Usually costs: 1 mark.
Mistake: Multiplying by the conjugate of the numerator
Mixing up which conjugate clears the denominator.
Do this: Always multiply top and bottom by the conjugate of the denominator.
Usually costs: 1-2 marks.
Mistake: Writing $∣ a + ib ∣ = a + b$ or $a^{2} - b^{2}$
Including i when squaring.
Do this: $∣ a + ib ∣ = a^{2} + b^{2}$ , using the real numbers a and b.
Usually costs: 1 mark.
Mistake: Comparing complex numbers with < or >
Carrying over the order of real numbers.
Do this: Only moduli (which are real) can be compared.
Usually costs: 1 mark in assertion-reason items.

NCERT exercise guide: Complex Numbers and Quadratic Equations

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 4.1 (algebra of complex numbers)

What it tests: Writing expressions in a + ib form, powers of i, cubes and fourth powers, multiplicative inverses, quotients of products.

How to approach it: Expand and use $i^{2} = - 1$ ; reduce powers mod 4; multiply by the conjugate to divide.

Same kind as Exercise 4.1

Find the multiplicative inverse of $3 - 2 i$ .

Show the solutionSolution

$∣ z ∣^{2} = 3 + 4 = 7$
$z^{- 1} = \frac{3 + 2 i}{7} = \frac{3}{7} + \frac{2}{7} i$

Answer: $\frac{3}{7} + \frac{2}{7} i$

Miscellaneous Exercise on Chapter 4

What it tests: Powers of i, proofs about real parts and moduli, standard forms of compound fractions, finding real x and y, moduli of expressions, results with $∣ β ∣ = 1$ .

How to approach it: Simplify step by step; use $z \overset{z}{ˉ} = ∣ z ∣^{2}$ and the modulus rules; equate parts.

Same kind as Miscellaneous Exercise

If $(x + i y)^{2} = u + i v$ , show that $u^{2} + v^{2} = (x^{2} + y^{2})^{2}$ .

Show the solutionSolution

$(x + i y)^{2} = x^{2} + 2 i x y + i^{2} y^{2} = (x^{2} - y^{2}) + i (2 x y)$ , so $u = x^{2} - y^{2}$ , $v = 2 x y$
$u^{2} + v^{2} = x^{4} - 2 x^{2} y^{2} + y^{4} + 4 x^{2} y^{2} = x^{4} + 2 x^{2} y^{2} + y^{4} = (x^{2} + y^{2})^{2}$

Answer: Proved.

Practice questions on Complex Numbers and Quadratic Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The multiplicative inverse of $2 i$ is:
(a) $- \frac{i}{2}$
(b) $\frac{i}{2}$
(c) $- 2 i$
(d) $\frac{1}{2}$
Hint
Product must be 1.
Answer
(a) $- \frac{i}{2}$ , since $2 i \cdot (- \frac{i}{2}) = - i^{2} = 1$
Q2 · 1 mark · MCQExam level
If $z = 3 - 4 i$ , then $∣ z ∣$ is:
(a) 7
(b) 1
(c) 5
(d) 25
Hint
Pythagoras.
Answer
(c) 5
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): $z \overset{z}{ˉ}$ is a non-negative real number for every complex z.
Reason (R): $(a + ib) (a - ib) = a^{2} + b^{2}$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Expand.
Answer
(a)
Q4 · 1 mark · Very short answerExam level
Express $(1 + i)^{6}$ in the form a + ib.
Hint
Square first, then cube.
Answer
$(1 + i)^{2} = 2 i$ , so $(1 + i)^{6} = (2 i)^{3} = 8 i^{3} = - 8 i = 0 - 8 i$ .
Q5 · 2 marks · Very short answerHarder exam level
Express $(3 - 2 i) (2 + 3 i)$ in the form a + ib and find its conjugate and modulus.
Hint
Use $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ to check.
Answer
$6 + 9 i - 4 i + 6 = 12 + 5 i$ ; conjugate $12 - 5 i$ ; modulus 13 (= $13 \times 13$ ).
Q6 · 2 marks · Very short answerHarder exam level
Find real x and y if $x + 4 y i = i x + y + 3$ .
Hint
Group real and imaginary parts on the right.
Answer
Right side $= (y + 3) + i x$ . So $x = y + 3$ and $4 y = x$ ; then $4 y = y + 3$ , y = 1, x = 4.
Q7 · 2 marks · Very short answerHarder exam level
Express $\frac{3 + 4 i}{2 - i} + \frac{1}{i}$ in the form a + ib.
Hint
Do each part separately.
Answer
$\frac{( 3 + 4 i ) ( 2 + i )}{5} = \frac{2 + 11 i}{5}$ ; $\frac{1}{i} = - i$ . Sum $= \frac{2}{5} + \frac{6}{5} i$ .
Q8 · 3 marks · Short answerHarder exam level
If $a + ib = \frac{c + i}{c - i}$ , where c is real, prove that $a^{2} + b^{2} = 1$ .
Hint
Take moduli.
Answer
$a^{2} + b^{2} = ∣ a + ib ∣^{2} = \frac{∣ c + i ∣ ^{2}}{∣ c - i ∣ ^{2}} = \frac{c ^{2} + 1}{c ^{2} + 1} = 1$ .
Q9 · 3 marks · Short answerHarder exam level
Show that $i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} = 0$ for every natural number n, and use it to find $\sum_{k = 1}^{100} i^{k}$ .
Hint
Factor out $i^{n}$ .
Answer
$i^{n} (1 + i + i^{2} + i^{3}) = i^{n} (1 + i - 1 - i) = 0$ . The 100 terms form 25 such blocks, so the sum is 0.
Q10 · 2 marks · Very short answerHOTS / competency
Find all complex numbers z with $z^{2} = \overset{z}{ˉ}$ that are purely real.
Hint
If z is real, $\overset{z}{ˉ} = z$ .
Answer
For real z, $z^{2} = z$ gives z = 0 or z = 1.
Q11 · 3 marks · Short answerHOTS / competency
If $∣ z_{1} ∣ = ∣ z_{2} ∣ = 1$ and $1 + z_{1} z_{2} \neq = 0$ , show that $\frac{z _{1} + z _{2}}{1 + z _{1} z _{2}}$ is a real number.
Hint
For a unit complex number, $\overset{z}{ˉ} = \frac{1}{z}$ ; show that w equals its own conjugate.
Answer
Let $w = \frac{z _{1} + z _{2}}{1 + z _{1} z _{2}}$ . Since $\overset{z_{1}}{ˉ} = \frac{1}{z _{1}}$ and $\overset{z_{2}}{ˉ} = \frac{1}{z _{2}}$ , $\overset{w}{ˉ} = \frac{\frac{1}{z _{1}} + \frac{1}{z _{2}}}{1 + \frac{1}{z _{1} z _{2}}} = \frac{( z _{1} + z _{2} ) / ( z _{1} z _{2} )}{( z _{1} z _{2} + 1 ) / ( z _{1} z _{2} )} = w$ . A number equal to its conjugate is real.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is $i^{4 k + 3}$ ?Show answer
−i.
2.What is $\frac{1}{i}$ ?Show answer
−i.
3.Write Re z and Im z for $z = - 5 + 2 i$ .Show answer
−5 and 2.
4.What is $z \overset{z}{ˉ}$ in terms of |z|?Show answer
$∣ z ∣^{2}$ .
5.Multiplicative inverse of $i$ ?Show answer
$- i$ , since $(i) (- i) = - i^{2} = 1$ .
6.What is $∣1 + i ∣$ ?Show answer
$2$ .
7.Where is $\overset{z}{ˉ}$ in the Argand plane?Show answer
The reflection of z in the real axis.
8.When is z purely imaginary in terms of $\overset{z}{ˉ}$ ?Show answer
When $\overset{z}{ˉ} = - z$ and $z \neq = 0$ .

Best order to study it

Understand why $i = - 1$ is needed and the form a + ib.
Learn equality of complex numbers and practise finding x and y.
Practise addition, subtraction and multiplication with $i^{2} = - 1$ .
Learn the cycle of powers of i, including negative powers.
Learn square roots of negative numbers and the multiplication trap.
Learn the conjugate, modulus and their properties, then the inverse.
Practise division by the conjugate method.
Finish with the Argand plane: plotting, distance and reflection.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you explain why $x^{2} + 1 = 0$ has no real root?
Can you write the real and imaginary parts of any complex number?
Can you solve for real x and y from a complex equation?
Can you expand $(a + ib)^{2}$ and $(a + ib)^{3}$ ?
Can you evaluate any power of i, positive or negative?
Can you simplify products of square roots of negative numbers correctly?
Can you find the conjugate, modulus and inverse of a complex number?
Can you divide two complex numbers and give a + ib?
Can you use $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ to find moduli quickly?
Can you plot a number and its conjugate in the Argand plane?

Complex Numbers and Quadratic Equations: frequently asked questions

Why does the chapter title mention quadratic equations if they are removed?

The NCERT chapter keeps its old title, but the section on solving quadratic equations was removed in the rationalisation. CBSE lists solving quadratics with real coefficients in the complex number system, and the polar form, as assessed only formatively for 2026-27, so they do not appear in the annual exam.

Is the polar (modulus-argument) form in the exam?

No. Polar representation is taught but assessed only formatively. The annual exam covers the algebra of complex numbers, conjugate, modulus, inverse and the Argand plane.

Is i a real number?

No. No real number squares to a negative number. i is a new number with i^2 = -1; every real number is a complex number, but not the other way round.

Can we say one complex number is greater than another?

No. Complex numbers cannot be arranged in order in a way that respects the rules of inequalities. You can compare their moduli, which are real numbers.

How is the Class 11 Maths exam set?

It is a school annual exam of 80 marks (3 hours) plus 20 marks of internal assessment; Unit II, Algebra, which includes this chapter, carries 25 marks. CBSE gives no Class 11 sample paper, so schools usually mix MCQs, assertion-reason, short and long answers and case-based questions.

Based on: CBSE Mathematics (041) syllabus 2026-27 (Classes XI-XII); NCERT Mathematics, Textbook for Class XI, Chapter 4 (Reprint 2026-27)