Complex Numbers and Quadratic Equations: Class 11 Maths · Chapter 4

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

60 min
1 simulation
27 practice questions
19 MCQs

What you will learn in Complex Numbers and Quadratic Equations

Explain why the real numbers must be extended to solve x² + 1 = 0, and describe a complex number a + ib by its real and imaginary parts.
Use the equality of complex numbers to find unknown real quantities by comparing real and imaginary parts.
Add, subtract, multiply and divide complex numbers, and state the laws these operations obey.
Simplify any integer power of i and write square roots of negative real numbers as multiples of i without misusing √a√b = √(ab).
Find the modulus, conjugate and multiplicative inverse of a complex number, and use their properties to simplify expressions and prove results.
Represent complex numbers in the Argand plane, reading the modulus as distance from the origin and the conjugate as a reflection in the real axis.

Original companion text aligned to NCERT Mathematics Class 11, Chapter 4, pp. 76–88. Not affiliated with or endorsed by NCERT.

4.1 Introduction

Type $\sqrt{- 9}$ into a phone calculator and it will most likely answer Error. Within the real numbers the calculator is right: the square of any real number is zero or positive, so no real number squares to $- 9$ . In this chapter we build a larger number system in which such square roots do exist, and we learn to calculate in it as freely as with ordinary numbers.

This is not the first time the number system has grown. Whenever a simple equation had no answer, mathematicians enlarged the system instead of giving up. Fig. 4.1 retraces the climb: $x + 5 = 2$ needs negative integers, $3 x = 2$ needs fractions and $x^{2} = 2$ needs irrational numbers. The equation $x^{2} + 1 = 0$ is the next rung.

5/5 $x^{2} = - 1$ has no real answer. The complex numbers supply $x = \pm i$ .

Fig. 4.1Each new kind of number answers an equation the old ones could not.

Five nested rounded boxes, the natural numbers innermost and the complex numbers outermost. Step 1 shows only the natural numbers 1, 2, 3 and so on. Step 2 adds the integers around them, with the equation x + 5 = 2 and its answer x = −3 on the right. Step 3 adds the rational numbers, with 3x = 2 and x = 2/3. Step 4 adds the real numbers, with x squared = 2 and x = ± root 2. Step 5 adds the complex numbers as the outermost box, with x squared = −1 and x = ± i.

There is a practical reason too. In Class 10 you solved $a x^{2} + b x + c = 0$ with the quadratic formula and had to stop whenever $b^{2} - 4 a c < 0$ . Once negative numbers have square roots, the formula works every time. Electrical engineers use the same numbers daily to describe alternating current in the power grid, and signal engineers use them to process the radio waves that carry phone calls.

4.2 Complex Numbers

We take the bold step directly. We introduce a new number $i$ , read iota, whose square is $- 1$ , and agree to keep every rule of algebra that still makes sense. This $i$ is the imaginary unit.

The imaginary unit

i = \sqrt{- 1}, i^{2} = - 1

(4.1)

The name imaginary is historical: these numbers are as well defined as any other. Combining $i$ with real numbers gives a whole new family, the complex numbers (सम्मिश्र संख्याएँ).

Definition

Complex number

A number $z = a + i b$ built from two real numbers $a$ and $b$ and the imaginary unit $i$ . We call $a$ the real part of $z$ ( $Re z$ ) and $b$ its imaginary part ( $Im z$ ).

ExampleFor $z = - 4 + \frac{3}{2} i$ : $Re z = - 4$ and $Im z = \frac{3}{2}$ .

Notice that both parts are real numbers. The imaginary part of $3 - 2 i$ is $- 2$ , not $- 2 i$ , while its real part is $3$ . Fig. 4.2 makes this explicit by writing the number as $3 + (- 2) i$ . You may write $a + i b$ or $a + b i$ ; with a surd, prefer $i \sqrt{3}$ to $\sqrt{3} i$ so that nobody reads it as $\sqrt{3 i}$ .

2/3 $Re z = 3$ and $Im z = - 2$ ; the $i$ belongs to neither.

Fig. 4.2

z = 3 - 2 i

3 + (- 2) i

: both parts are real numbers.NCERT 4.2 · p. 76

The number z = 3 + (−2)i is written in large type. In step 2 the 3 is boxed and labelled real part, Re z = 3, and the −2 in brackets is boxed with a dashed outline and labelled imaginary part, Im z = −2; the letter i stays outside both boxes. In step 3 two chips appear underneath: 5 + 0i, labelled real, and 0 + 4i, labelled purely imaginary.

Two kinds of complex number deserve names. When $b = 0$ , the number $a + 0 i$ is just the real number $a$ , so every real number is also a complex number. When $a = 0$ and $b \neq 0$ , the number $0 + i b = i b$ is called purely imaginary, such as $7 i$ or $- \frac{1}{2} i$ . The set of all complex numbers is written $𝐂$ , and $𝐑 \subset 𝐂$ .

When are two complex numbers the same? Exactly when they match part by part. This rule turns a single equation between complex numbers into two equations between real numbers, a trick we will use again and again.There is no sensible way to say that one complex number is bigger than another. Statements such as $3 + i > 2$ are meaningless; only moduli, which are real, can be compared.

Equality

\begin{array}{c} a + i b = c + i d \\ ⟺ a = c and b = d \end{array}

(4.2)

where

$a, b, c, d$: real numbers

Example 4.1NCERT Example 1 · p. 77

Comparing parts

Find the real numbers $x$ and $y$ for which $(3 x + 2 y) + i (x - y) = 12 - i .$

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

What is

Im (5 - 7 i)

- 7 i

7

- 7

5

Show answer

Answer (C) $- 7$

The imaginary part is the real number that multiplies

i

, sign included:

- 7

. Writing

- 7 i

is the most common slip;

Im z

never contains

i

4.3 Algebra of Complex Numbers

The working rule for this whole section is short: calculate with $i$ exactly as you would with a letter such as $x$ , and whenever $i^{2}$ appears, replace it by $- 1$ . Every operation below then returns a number of the form $a + i b$ again, and the familiar laws of arithmetic survive.

4.3.1 Addition of two complex numbers

To add two complex numbers, add the real parts and add the imaginary parts, just as you collect like terms in $(4 - 3 x) + (- 1 + 7 x) .$

Sum

\begin{array}{c} (a + i b) + (c + i d) \\ = (a + c) + i (b + d) \end{array}

(4.3)

For example, $(4 - 3 i) + (- 1 + 7 i) = 3 + 4 i .$ Since each part is added as an ordinary real number, addition keeps the laws you know: it is closed (the sum is again complex), commutative and associative. The number $0 = 0 + 0 i$ is the additive identity, because $z + 0 = z$ , and each $z = a + i b$ has an additive inverse $- z = - a - i b$ with $z + (- z) = 0$ . Table 4.1 lists these laws next to those for multiplication.

4.3.2 Difference of two complex numbers

Subtraction is defined through the additive inverse: $z_{1} - z_{2} = z_{1} + (- z_{2}) .$ In practice you subtract the real parts and subtract the imaginary parts.

Is subtraction commutative? Test it with $z_{1} = 5 + 2 i$ and $z_{2} = 3 - 4 i$ . Adding $- z_{2} = - 3 + 4 i$ gives $z_{1} - z_{2} = 2 + 6 i,$ but in the other order $z_{2} - z_{1} = - 2 - 6 i .$ The answers are negatives of each other, so subtraction is not commutative, just as with real numbers.

4.3.3 Multiplication of two complex numbers

Multiply two complex numbers as you would multiply two binomials, then tidy up with $i^{2} = - 1$ . Expanding $(a + i b) (c + i d)$ produces four terms, one for each cell of Fig. 4.3.

Product

\begin{aligned} (a + i b) (c + i d) \\ = (a c - b d) \\ + i (a d + b c) \end{aligned}

(4.4)

4/4Real cells give $a c - b d$ ; imaginary cells give $a d + b c$ .

Fig. 4.3Four products, and the one with

i^{2}

changes sign.NCERT 4.3.3 · p. 78

A grid with c and + id across the top and a and + ib down the side. Step 1 shows the empty grid. Step 2 fills the cells with ac, iad, ibc and i squared bd. Step 3 points an arrow labelled i squared = −1 at the last cell, which now reads −bd. Step 4 shades the cells ac and −bd as real and the cells iad and ibc, with dashed borders, as imaginary, and writes the result (ac − bd) + i(ad + bc) underneath.

With numbers: $(4 + 3 i) (2 - 5 i) = 8 - 20 i + 6 i - 15 i^{2} .$ The last term is $- 15 i^{2} = + 15$ , so the product is $23 - 14 i$ . There is no need to memorise (4.4): expand, then replace $i^{2}$ .

Table 4.1Laws of addition and multiplication for complex numbers u, v, w

Law	For addition	For multiplication
Closure	$u + v$ is complex	$u v$ is complex
Commutative	$u + v = v + u$	$u v = v u$
Associative	$(u + v) + w = u + (v + w)$	$(u v) w = u (v w)$
Identity	$0 = 0 + 0 i$ , $u + 0 = u$	$1 = 1 + 0 i$ , $u \cdot 1 = u$
Inverse	$- u$ , with $u + (- u) = 0$	$u^{- 1}$ for $u \neq 0$ , with $u u^{- 1} = 1$

The two operations are tied together by the distributive law: $z_{1} (z_{2} + z_{3}) = z_{1} z_{2} + z_{1} z_{3}$ and $(z_{1} + z_{2}) z_{3} = z_{1} z_{3} + z_{2} z_{3} .$ Each law can be checked by writing $z_{1}, z_{2}, z_{3}$ as $a + i b$ and comparing parts; from now on we use them freely. Only the multiplicative inverse needs real work.

DerivationFinding the reciprocal

Every complex number except $0$ has a multiplicative inverse. For $z = a + i b \neq 0$ we look for real $x, y$ with $(a + i b) (x + i y) = 1 + 0 i .$ Expanding with (4.4) and comparing parts gives two linear equations.

\begin{aligned} a x - b y & = 1 \\ b x + a y & = 0 \end{aligned}

Multiply the first equation by $a$ , the second by $b$ , and add: $(a^{2} + b^{2}) x = a$ . Multiply the first by $- b$ , the second by $a$ , and add: $(a^{2} + b^{2}) y = - b$ . As $z \neq 0$ , the numbers $a$ and $b$ are not both zero, so $a^{2} + b^{2} > 0$ and we may divide.

Multiplicative inverse

\begin{aligned} z^{- 1} & = \frac{a}{a^{2} + b^{2}} \\ + i \frac{- b}{a^{2} + b^{2}} \end{aligned}

(4.5)

It is also written $\frac{1}{z}$ . Multiplying out $z \cdot z^{- 1}$ confirms the answer is $1$ .

4.3.4 Division of two complex numbers

Division is multiplication by an inverse. For $z_{2} \neq 0$ we define $\frac{z_{1}}{z_{2}} = z_{1} \cdot \frac{1}{z_{2}},$ and (4.5) supplies $\frac{1}{z_{2}}$ .

Example

One complex number divided by another

Divide $10 + 5 i$ by $1 + 2 i$ .

Try it yourself first. Then reveal the solution one step at a time.

Section 4.4 gives a quicker route that needs no formula: multiply the top and the bottom by the conjugate of the bottom.

4.3.5 Power of i

Multiplying by $i$ again and again produces only four values. From $i^{2} = - 1$ we get $i^{3} = i^{2} \cdot i = - i$ and $i^{4} = (i^{2})^{2} = 1$ . Then $i^{5} = i^{4} \cdot i = i$ , and the cycle starts again (Fig. 4.4).

Negative powers follow from $i \cdot (- i) = - i^{2} = 1,$ which says $i^{- 1} = - i$ . Hence $i^{- 2} = (- i)^{2} = - 1,$ $i^{- 3} = i^{- 2} \cdot i^{- 1} = i$ and $i^{- 4} = 1$ . Going backwards visits the same four values in reverse order.

Powers of i

\begin{aligned} i^{4 k} & = 1 \\ i^{4 k + 1} & = i \\ i^{4 k + 2} & = - 1 \\ i^{4 k + 3} & = - i \end{aligned}

(4.6)

where

$k$: any integer, positive, negative or zero

3/4 $i^{3} = - i$ and $i^{4} = 1$ : the exponent's remainder on division by $4$ picks the station.

Fig. 4.4Each multiplication by

i

moves one station anticlockwise; four moves bring you home.NCERT 4.3.5 · p. 79

A ring carries four stations: 1 on the right, i at the top, −1 on the left and −i at the bottom. Step 1 draws an arrow labelled times i from 1 to i, with i to the power 1 = i in the centre. Step 2 adds the arrow from i to −1, with i squared = −1. Step 3 completes the loop through −i back to 1, shows i to the power 4 = 1, and labels the stations with the exponents 4k, 4k + 1, 4k + 2 and 4k + 3. Step 4 fades the loop and draws a dashed arrow labelled divide by i going backwards from 1 to −i, with i to the power −1 = −i.

So to simplify $i^{n}$ , divide $n$ by $4$ and keep the remainder $r \in {0, 1, 2, 3}$ ; then $i^{n} = i^{r}$ . The rule works for negative $n$ as long as the remainder is taken between $0$ and $3$ , for instance $- 22 = 4 (- 6) + 2$ .

Example

Taming large powers

Simplify (a) $i^{101}$ , (b) $i^{- 22}$ and (c) $i + i^{2} + i^{3} + i^{4}$ .

Try it yourself first. Then reveal the solution one step at a time.

4.3.6 The square roots of a negative real number

Since $i^{2} = - 1$ and also $(- i)^{2} = (- 1)^{2} i^{2} = - 1,$ the number $- 1$ has two square roots, $i$ and $- i$ . The symbol $\sqrt{- 1}$ is reserved for one of them, $i$ , just as $\sqrt{9}$ means $3$ and not $- 3$ .

The same holds for every negative number. If $a > 0$ , then $(\pm i \sqrt{a})^{2} = a i^{2} = - a,$ so $- a$ has the two square roots $\pm i \sqrt{a}$ , and the symbol $\sqrt{- a}$ always denotes $i \sqrt{a}$ . Thus $\sqrt{- 49} = 7 i$ and $\sqrt{- 5} = i \sqrt{5}$ .

Now a trap that catches many students (Fig. 4.5). The familiar rule $\sqrt{a} \sqrt{b} = \sqrt{a b}$ breaks down when $a$ and $b$ are both negative. In every other case it is safe: both numbers non-negative, or one positive and one negative (and if either is $0$ , both sides are $0$ ).

3/3Writing $\sqrt{(- 4) (- 9)} = \sqrt{36} = 6$ gives the wrong sign.

Fig. 4.5Change

\sqrt{- a}

i \sqrt{a}

before multiplying.NCERT 4.3.6 · p. 80

At the top is the product root of −4 times root of −9. Step 1 shows only the question. Step 2 opens a solid green box on the left headed roots first: 2i times 3i, then 6 i squared, then −6, marked with a tick and the word correct. Step 3 opens a dashed box on the right headed merge first: root of (−4)(−9), then root 36, then 6, crossed out and marked wrong.

No clever fix can rescue the rule. If it held for $a = b = - 1$ , we would get $i^{2} = \sqrt{- 1} \sqrt{- 1} = \sqrt{(- 1) (- 1)} = \sqrt{1} = 1,$ contradicting $i^{2} = - 1$ . The safe habit is to rewrite every $\sqrt{- a}$ as $i \sqrt{a}$ before doing anything else.

4.3.7 Identities

The algebraic identities you know for real numbers were proved using only the commutative, associative and distributive laws. Complex numbers obey the same laws, so the same identities hold for all complex $z_{1}$ and $z_{2}$ . Here is one proof with every law named.

TheoremDifference of squares

For all complex numbers $z_{1}$ and $z_{2}$ , $z_{1}^{2} - z_{2}^{2} = (z_{1} + z_{2}) (z_{1} - z_{2}) .$

Squares and cubes

\begin{aligned} (z_{1} \pm z_{2})^{2} \\ = z_{1}^{2} \pm 2 z_{1} z_{2} + z_{2}^{2} \\ (z_{1} \pm z_{2})^{3} \\ = z_{1}^{3} \pm 3 z_{1}^{2} z_{2} \\ + 3 z_{1} z_{2}^{2} \pm z_{2}^{3} \end{aligned}

(4.7)

Expanding a cube with (4.7) is faster and safer than multiplying three brackets, as Example 4.3 shows. Before that, two warm-up products.

Example 4.2NCERT Example 2 · p. 80

Products of imaginary numbers

Write in the form $a + i b$ : (i) $(6 i) (- \frac{2}{3} i)$ (ii) $(- 2 i)^{3} {(\frac{1}{4} i)}^{2}$

Try it yourself first. Then reveal the solution one step at a time.

Example 4.3NCERT Example 3 · p. 81

Cube of a binomial

Express $(2 + 3 i)^{3}$ in the form $a + i b$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 4.4NCERT Example 4 · p. 81

Square roots of negatives in a product

Express $(\sqrt{5} + \sqrt{- 3}) (\sqrt{5} - 2 \sqrt{- 3})$ in the form $a + i b$ .

Try it yourself first. Then reveal the solution one step at a time.

4.4 The Modulus and the Conjugate of a Complex Number

Two numbers attached to every complex number make division painless and, in Section 4.5, give complex numbers a geometric meaning.

Definition

Modulus

The modulus of $z = a + i b$ , written $∣ z ∣$ , is $\sqrt{a^{2} + b^{2}}$ : a real number that is never negative.

Example $∣ 5 - 12 i ∣ = \sqrt{25 + 144} = 13$

Definition

Conjugate

The conjugate of $z = a + i b$ is $\overset{ˉ}{z} = a - i b$ : keep the real part and reverse the sign of the imaginary part.

Example $\overline{4 + 7 i} = 4 - 7 i$

Modulus and conjugate

\begin{array}{c} ∣ z ∣ = \sqrt{a^{2} + b^{2}} \\ \overset{ˉ}{z} = a - i b \end{array}

(4.8)

A few more: $∣ - 2 + i ∣ = \sqrt{5}$ , $\overline{- 3 i} = 3 i$ and $\overline{6} = 6$ , because a real number is its own conjugate. For a real number $a$ , $∣ a + 0 i ∣ = \sqrt{a^{2}} = ∣ a ∣,$ so the modulus (मापांक) extends the absolute value you already know. Taking the conjugate twice returns the original number: $\overline{\overline{z}} = z$ .

A number and its conjugate make a useful pair (Fig. 4.6). Their sum $2 a$ is real, their difference $2 i b$ is purely imaginary, and their product is real and never negative.

4/4 $z \overset{ˉ}{z} = a^{2} - (i b)^{2} = a^{2} + b^{2},$ which is $∣ z ∣^{2}$ .

Fig. 4.6Adding, subtracting and multiplying a number and its conjugate.NCERT 4.4 · p. 81

A card on the left shows z = a + ib above z-bar = a − ib. Step 1 shows only the pair. Step 2 strikes out the terms + ib and − ib, and a row on the right reads z + z-bar = 2a, labelled real. Step 3 strikes out the two a terms instead, and a second row reads z − z-bar = 2ib, labelled purely imaginary. Step 4 removes the strike marks and adds a third row, z times z-bar = a squared + b squared, labelled real, never negative.

A number times its conjugate

\begin{array}{c} z \overset{ˉ}{z} = (a + i b) (a - i b) \\ = a^{2} + b^{2} = ∣ z ∣^{2} \end{array}

(4.9)

Compare (4.9) with the reciprocal (4.5): its numerator is $a - i b = \overset{ˉ}{z}$ and its denominator is $a^{2} + b^{2} = ∣ z ∣^{2}$ . That gives the inverse a neat form. It also gives division a neat method: multiply the numerator and the denominator by the conjugate of the denominator, which turns the denominator into the real number $∣ z_{2} ∣^{2}$ .

Inverse and quotient

\begin{array}{c} z^{- 1} = \frac{\overset{ˉ}{z}}{∣ z ∣^{2}} \\ \frac{z_{1}}{z_{2}} = \frac{z_{1} {\overset{ˉ}{z}}_{2}}{∣ z_{2} ∣^{2}} \end{array}

(4.10)

where

$z, z_{2}$: non-zero complex numbers

PropertyRules for modulus and conjugate

For complex numbers $z_{1}$ and $z_{2}$ : (i) $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣ ∣ z_{2} ∣;$ (ii) $∣ \frac{z_{1}}{z_{2}} ∣ = \frac{∣ z_{1} ∣}{∣ z_{2} ∣}$ if $z_{2} \neq 0$ ; (iii) $\overline{z_{1} z_{2}} = {\overset{ˉ}{z}}_{1} {\overset{ˉ}{z}}_{2};$ (iv) $\overline{z_{1} \pm z_{2}} = {\overset{ˉ}{z}}_{1} \pm {\overset{ˉ}{z}}_{2};$ (v) $\overline{(\frac{z_{1}}{z_{2}})} = \frac{{\overset{ˉ}{z}}_{1}}{{\overset{ˉ}{z}}_{2}}$ if $z_{2} \neq 0$ .

Put $z_{1} = a + i b$ , $z_{2} = c + i d$ . For (iv), both sides equal $(a \pm c) - i (b \pm d) .$ For (iii), conjugate the product formula (4.4): the real part $a c - b d$ stays and the imaginary part $a d + b c$ changes sign. Multiplying ${\overset{ˉ}{z}}_{1} = a - i b$ by ${\overset{ˉ}{z}}_{2} = c - i d$ directly gives the same real part $a c - b d$ and the imaginary part $- (a d + b c)$ , so the two agree.

For (i), use (4.9) and then (iii): $∣ z_{1} z_{2} ∣^{2} = (z_{1} z_{2}) \overline{z_{1} z_{2}} = (z_{1} {\overset{ˉ}{z}}_{1}) (z_{2} {\overset{ˉ}{z}}_{2}) = ∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2} .$ Both moduli are non-negative, so take square roots. Finally, write $z_{1} = \frac{z_{1}}{z_{2}} \cdot z_{2}$ and apply (i) and (iii) to this product; dividing by $∣ z_{2} ∣$ or by ${\overset{ˉ}{z}}_{2}$ gives (ii) and (v).

Notice what is missing from the list: the modulus does not add. For $z_{1} = 1$ and $z_{2} = - 1$ we get $∣ z_{1} + z_{2} ∣ = 0$ but $∣ z_{1} ∣ + ∣ z_{2} ∣ = 2$ . In general only $∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣$ holds, the triangle inequality, which you can see in the sum mode of Fig. 4.11.

Example 4.5NCERT Example 5 · p. 82

A multiplicative inverse

Find the multiplicative inverse of $- 3 + 4 i$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 4.6NCERT Example 6 · p. 82

Dividing with the conjugate

Express in the form $a + i b$ : (i) $\frac{3 + i \sqrt{3}}{1 - i \sqrt{3}}$ (ii) $i^{- 45}$

Try it yourself first. Then reveal the solution one step at a time.

Practice · 8 questions

Practice 4.1: questions with answers

Write every answer in the form $a + i b$ unless the question asks for something else.

1
Short answerEasyNCERT Exercise 4.1, Q1–Q3
Express in the form $a + i b$ .
1. $(- 4 i) (\frac{3}{8} i)$
  Show solution
  Answer $\frac{3}{2} + 0 i$
  $(- 4 i) (\frac{3}{8} i) = - \frac{12}{8} i^{2} = \frac{3}{2} .$
2. $i^{14} + i^{27}$
  Show solution
  Answer $- 1 - i$
  $14 = 4 \cdot 3 + 2$ and $27 = 4 \cdot 6 + 3$ , so the sum is $i^{2} + i^{3} = - 1 - i .$
3. $i^{- 71}$
  Show solution
  Answer $0 + i$
  $- 71 = 4 (- 18) + 1,$ so $i^{- 71} = i$ . Check: $i^{71} = i^{3} = - i$ and $\frac{1}{- i} = i$ .
2
Short answerEasyNCERT Exercise 4.1, Q4–Q7
Simplify and write in the form $a + i b$ .
1. $2 (5 - 3 i) - i (5 - 3 i)$
  Show solution
  Answer $7 - 11 i$
  $10 - 6 i - 5 i + 3 i^{2} = 10 - 3 - 11 i = 7 - 11 i .$
2. $(- 2 + 5 i) - (3 - 4 i)$
  Show solution
  Answer $- 5 + 9 i$
  Subtract part by part: $(- 2 - 3) + (5 + 4) i = - 5 + 9 i .$
3. $(\frac{3}{4} - \frac{1}{2} i) - (2 + \frac{5}{3} i)$
  Show solution
  Answer $- \frac{5}{4} - \frac{13}{6} i$
  Real part: $\frac{3}{4} - 2 = - \frac{5}{4}$ . Imaginary part: $- \frac{1}{2} - \frac{5}{3} = - \frac{13}{6} .$
4. $[(\frac{1}{2} + \frac{3}{2} i) + (3 - \frac{1}{2} i)] - (- \frac{5}{2} + 2 i)$
  Show solution
  Answer $6 - i$
  The square bracket is $\frac{7}{2} + i$ . Subtracting gives $(\frac{7}{2} + \frac{5}{2}) + (1 - 2) i = 6 - i .$
3
Short answerMediumNCERT Exercise 4.1, Q8–Q10
Expand and write in the form $a + i b$ .
1. $(1 + i)^{6}$
  Show solution
  Answer $0 - 8 i$
  $(1 + i)^{2} = 1 + 2 i + i^{2} = 2 i,$ so $(1 + i)^{6} = (2 i)^{3} = 8 i^{3} = - 8 i .$
2. ${(\frac{1}{2} - 2 i)}^{3}$
  Show solution
  Answer $- \frac{47}{8} + \frac{13}{2} i$
  Use $(z_{1} - z_{2})^{3}$ from (4.7) with $z_{1} = \frac{1}{2}$ and $z_{2} = 2 i$ ; note $(2 i)^{2} = - 4$ and $(2 i)^{3} = - 8 i$ .
  $\begin{aligned} \frac{1}{8} - \frac{3}{4} (2 i) + \frac{3}{2} (2 i)^{2} - (2 i)^{3} \\ = \frac{1}{8} - \frac{3}{2} i - 6 + 8 i \\ = - \frac{47}{8} + \frac{13}{2} i \end{aligned}$
3. ${(- 1 + \frac{1}{2} i)}^{3}$
  Show solution
  Answer $- \frac{1}{4} + \frac{11}{8} i$
  Square first: ${(- 1 + \frac{1}{2} i)}^{2} = 1 - i - \frac{1}{4} = \frac{3}{4} - i .$
  $\begin{aligned} (\frac{3}{4} - i) (- 1 + \frac{1}{2} i) \\ = - \frac{3}{4} + \frac{3}{8} i + i + \frac{1}{2} \\ = - \frac{1}{4} + \frac{11}{8} i \end{aligned}$
4
Short answerEasyNCERT Exercise 4.1, Q11–Q13
Find the multiplicative inverse of each number.
1. $3 - 5 i$
  Show solution
  Answer $\frac{3}{34} + \frac{5}{34} i$
  $∣ z ∣^{2} = 9 + 25 = 34$ and $\overset{ˉ}{z} = 3 + 5 i$ , so $z^{- 1} = \frac{3 + 5 i}{34} .$
2. $\sqrt{7} - 3 i$
  Show solution
  Answer $\frac{\sqrt{7}}{16} + \frac{3}{16} i$
  $∣ z ∣^{2} = 7 + 9 = 16$ , so $z^{- 1} = \frac{\sqrt{7} + 3 i}{16} .$
3. $2 i$
  Show solution
  Answer $0 - \frac{1}{2} i$
  $\overset{ˉ}{z} = - 2 i$ and $∣ z ∣^{2} = 4$ , so $z^{- 1} = - \frac{1}{2} i$ . Check: $2 i \cdot (- \frac{1}{2} i) = - i^{2} = 1.$
Short answerHardNCERT Exercise 4.1, Q14
Express $\frac{(\sqrt{3} + 2 i) (\sqrt{3} - 2 i)}{(1 + i \sqrt{2})^{2}}$ in the form $a + i b$ .
Show solution
Answer $- \frac{7}{9} - \frac{14 \sqrt{2}}{9} i$
Numerator: $∣ \sqrt{3} + 2 i ∣^{2} = 3 + 4 = 7.$ Denominator: $(1 + i \sqrt{2})^{2} = 1 + 2 i \sqrt{2} - 2 = - 1 + 2 \sqrt{2} i .$
Multiply the top and the bottom by $- 1 - 2 \sqrt{2} i$ ; the new denominator is $1 + 8 = 9$ .
$\begin{array}{c} \frac{7 (- 1 - 2 \sqrt{2} i)}{9} \\ = - \frac{7}{9} - \frac{14 \sqrt{2}}{9} i \end{array}$
Quick answerEasy
$i^{2026}$ equals
$1$ $- 1$ $i$ $- i$
Show answer
Answer (B) $- 1$
$2026 = 4 \times 506 + 2,$ so $i^{2026} = i^{2} = - 1$ .
Quick answerEasy
$\frac{1}{1 - i}$ equals
$1 + i$ $\frac{1}{2} - \frac{1}{2} i$ $\frac{1}{2} + \frac{1}{2} i$ $- 1 + i$
Show answer
Answer (C) $\frac{1}{2} + \frac{1}{2} i$
$\frac{1 + i}{(1 - i) (1 + i)} = \frac{1 + i}{2} .$ Option B is the conjugate of the answer.
Quick answerMedium
Find $Re (\frac{5}{2 - i}) .$
Show answer
Answer 2
$\frac{5}{2 - i} = \frac{5 (2 + i)}{4 + 1} = 2 + i,$ so the real part is $2$ .

4.5 Argand Plane and Polar Representation

A complex number $a + i b$ is fixed completely by the ordered pair of real numbers $(a, b)$ , and an ordered pair is exactly what names a point of the coordinate plane. So we let the point $(a, b)$ stand for the number $a + i b$ . Every complex number gets its own point, and every point its own complex number.

Such a plane goes by two names, the complex plane and the Argand plane (आर्गंड तल), the second after the Swiss-born amateur mathematician Jean-Robert Argand, who described it in 1806.The Norwegian-Danish surveyor Caspar Wessel had published the same idea in 1799, but his paper went almost unnoticed for a century. The horizontal axis carries the numbers $a + 0 i$ and is called the real axis; the vertical axis carries the numbers $0 + i b$ and is the imaginary axis (Fig. 4.7).

3/3 $a + i b$ is found by going $a$ across and $b$ up, like a coordinate pair.

Fig. 4.7The point

(a, b)

stands for the complex number

a + i b

.NCERT Fig. 4.1 · p. 83

A grid with a horizontal axis labelled Re, the real axis, and a vertical axis labelled Im, the imaginary axis, crossing at O. Step 1 marks the real numbers −3 and 4 as round dots on the real axis. Step 2 adds the purely imaginary numbers 2i and −3i as square dots on the imaginary axis. Step 3 adds four diamond-shaped points, 2 + 3i in the first quadrant, −4 + 2i in the second, −3 − 2i in the third and 4 − 2i in the fourth, with dashed lines from 2 + 3i down to the real axis and across to the imaginary axis.

Reading a point is like reading coordinates: walk $a$ units along the real axis, then $b$ units parallel to the imaginary axis. The number $2 + 3 i$ is the point $(2, 3)$ in the first quadrant, while $- 3 - 2 i$ sits at $(- 3, - 2)$ in the third. The axes themselves hold the real numbers and the purely imaginary numbers.

The plane gives the modulus a meaning. The point for $z = a + i b$ is a horizontal distance $∣ a ∣$ and a vertical distance $∣ b ∣$ from the origin O, so by Pythagoras its distance from O is $\sqrt{a^{2} + b^{2}} = ∣ z ∣$ (Fig. 4.8). Many numbers share one modulus: all the numbers with $∣ z ∣ = 5$ lie on the circle of radius $5$ about O.

2/3The legs are $4$ and $3$ , so $∣ z ∣ = \sqrt{4^{2} + 3^{2}} = 5.$

Fig. 4.8

∣ z ∣

is the distance of the point

z

from O.NCERT Fig. 4.2 · p. 84

An Argand plane with the point z = 4 + 3i. Step 2 draws dashed legs of lengths 4 along the real axis and 3 upwards, with a right-angle mark, the segment from O to z labelled |z| = 5, and the working root of 4 squared plus 3 squared = 5. Step 3 adds a dotted circle of radius 5 about O passing through −3 + 4i, −5, −5i and 3 − 4i, all marked as points.

The conjugate becomes a reflection. The points $(a, b)$ for $z$ and $(a, - b)$ for $\overset{ˉ}{z}$ lie on one vertical line, the same distance above and below the real axis. So $\overset{ˉ}{z}$ is the mirror image of $z$ in the real axis (Fig. 4.9). In the same way, $- z$ is the image of $z$ under a half-turn about O, and $- \overset{ˉ}{z}$ is its mirror image in the imaginary axis.

3/3 $- z$ is the half-turn image through O; $- \overset{ˉ}{z}$ is the mirror image in the imaginary axis.

Fig. 4.9

\overset{ˉ}{z}

is the mirror image of

z

in the real axis.NCERT Fig. 4.3 · p. 84

An Argand plane with the point z = 3 + 2i. Step 2 highlights the real axis, draws a dashed vertical segment from z down to the conjugate 3 − 2i with a right-angle mark where it crosses the axis and equal tick marks on its two halves. Step 3 adds the points −z = −3 − 2i, joined to z by a dashed line through O, and −z-bar = −3 + 2i, and a dotted rectangle through all four points.

Example

Distances on the Argand plane

For $z = - 4 + 3 i$ , find $∣ z ∣$ , plot $\overset{ˉ}{z}$ and $- z$ , and find the distance from $z$ to each of them.

Try it yourself first. Then reveal the solution one step at a time.

SimulationModulus and conjugate on the plane

A square grid from −6 to 6 on both axes with a fixed point O at the origin and a draggable point z that starts at (3, 2), that is 3 + 2i. The segment from O to z is drawn and its length is measured, which is the modulus of z. The mirror image of each point in the horizontal axis is also drawn, which for z is its conjugate. The coordinates update as z is dragged between whole-number points.

Fig. 4.10Read the horizontal axis as the real axis and the vertical axis as the imaginary axis. Drag

z

: the measured length is

∣ z ∣

and the mirror image is

\overset{ˉ}{z}

Try this

Drag $z$ to $(4, 3)$ and read the distance. Then find three more grid points at the same distance from O.
Put $z$ on the horizontal axis. Where is its mirror image, and what does that say about the conjugate of a real number?
Move $z$ into the third quadrant. Which quadrant holds $\overset{ˉ}{z}$ ?

InteractiveComplex arithmetic lab

A square grid from −8 to 8 on both axes, the real axis horizontal, with two arrows from the origin: z1, solid blue, starting at 2 + i, and z2, dashed orange, starting at 1 + 2i. Their tips can be dragged to whole-number points from −5 to 5 or set with four number boxes. Six buttons choose the view. Sum shades the parallelogram on z1 and z2 and draws the sum as its diagonal. Difference draws minus z2 from the tip of z1. Product draws z1 times z2 and compares the moduli. Quotient multiplies by the conjugate and gives the answer as a fraction and a decimal. Conjugate reflects z1 in the real axis and draws the circle of radius |z1|. Times i shows z1, i z1, i squared z1 and i cubed z1 at the corners of a square. A panel states every result in words and symbols. Nothing moves until the learner acts.

Fig. 4.11Drag

z_{1}

and

z_{2}

or type their parts, then pick an operation. The panel shows the algebra and the grid shows where the answer lies.

Quick check

1 of 3

In which quadrant of the Argand plane does

- 3 - 4 i

lie?

FirstSecondThirdFourth

Show answer

Answer (C) Third

The real part is negative (left of the imaginary axis) and the imaginary part is negative (below the real axis), so the point is in the third quadrant. Choosing the second quadrant comes from checking only the real part.

Miscellaneous Examples

Example 4.7NCERT Example 7 · p. 85

Conjugate of a quotient

Find the conjugate of $\frac{(1 + 3 i) (2 - i)}{(1 - i) (3 + i)} .$

Try it yourself first. Then reveal the solution one step at a time.

Example 4.8NCERT Example 8 · p. 85

A condition that forces a circle

Let $z = x + i y$ with $z \neq - 1$ . If $\frac{z - 1}{z + 1}$ is purely imaginary, prove that $x^{2} + y^{2} = 1$ .

Try it yourself first. Then reveal the solution one step at a time.

Geometrically, $∣ z ∣ = 1$ says that $z$ lies on the circle of radius $1$ about O, the unit circle (Fig. 4.12). The figure shows something more: the segments joining $z$ to $- 1$ and to $1$ always meet at a right angle, the angle in a semicircle. When you meet the argument of a complex number later, you will see that this right angle is exactly what makes the quotient purely imaginary.

3/3The two segments meet at $90^{\circ}$ : the angle in a semicircle.

Fig. 4.12Points with

∣ z ∣ = 1

make up the unit circle.NCERT Example 8 · p. 85

A circle of radius 1 centred at the origin O, crossing the real axis at −1 and 1. Step 1 shows the circle, labelled x squared + y squared = 1, and a dashed radius of length 1 to the point z = 0.6 + 0.8i. Step 2 joins z to −1 with a solid segment and to 1 with a dashed segment. Step 3 fades the radius and marks a right angle at z between the two segments, labelled 90 degrees.

Recap

Complex Numbers and Quadratic Equations: key points and summary

Real numbers cannot solve $x^{2} + 1 = 0$ ; adjoining $i$ with $i^{2} = - 1$ extends them to the complex numbers, and $𝐑 \subset 𝐂$ (Section 4.1).
A complex number is $z = a + i b$ with real part $a$ and imaginary part $b$ , both real; two complex numbers are equal when both parts agree (Section 4.2).
$(a + i b) \pm (c + i d) = (a \pm c) + i (b \pm d)$ (Section 4.3.1).
$(a + i b) (c + i d) = (a c - b d) + i (a d + b c) .$ Addition and multiplication are closed, commutative and associative, have identities $0$ and $1$ and inverses, and obey the distributive law (Section 4.3.3).
Every $z = a + i b \neq 0$ has the inverse $z^{- 1} = \frac{a - i b}{a^{2} + b^{2}},$ and $\frac{z_{1}}{z_{2}} = z_{1} \cdot \frac{1}{z_{2}}$ (Section 4.3.4).
$i^{4 k} = 1$ , $i^{4 k + 1} = i$ , $i^{4 k + 2} = - 1$ and $i^{4 k + 3} = - i$ for every integer $k$ (Section 4.3.5).
For $a > 0$ , $\sqrt{- a} = i \sqrt{a}$ ; never merge $\sqrt{a} \sqrt{b}$ into $\sqrt{a b}$ when $a$ and $b$ are both negative (Section 4.3.6).
Identities for real numbers, such as $(z_{1} + z_{2})^{2} = z_{1}^{2} + 2 z_{1} z_{2} + z_{2}^{2},$ hold for complex numbers too (Section 4.3.7).
$∣ z ∣ = \sqrt{a^{2} + b^{2}}$ , $\overset{ˉ}{z} = a - i b$ , $z \overset{ˉ}{z} = ∣ z ∣^{2}$ and $z^{- 1} = \overset{ˉ}{z} / ∣ z ∣^{2}$ ; moduli multiply and divide, and conjugates pass through all four operations (Section 4.4).
In the Argand plane $z$ is the point $(a, b)$ , $∣ z ∣$ is its distance from O and $\overset{ˉ}{z}$ is its mirror image in the real axis (Section 4.5).

Timeline

How impossible roots became numbers

c. 850 CE
Mahavira
In Ganita Sara Sangraha he notes that no number squares to a negative one, so negatives have no square roots among the known numbers.
1150
Bhaskara II
His Bijaganita makes the same point about negative quantities.
1545
Gerolamo Cardano
Splitting $10$ into two parts with product $40$ leads him to $5 \pm \sqrt{- 15}$ , which he dismisses as useless.
1629
Albert Girard
He argues that admitting such roots lets an equation have as many roots as its degree.
1777
Leonhard Euler
He introduces the letter $i$ for $\sqrt{- 1}$ .
1799–1806
Wessel and Argand
Each pictures complex numbers as points of a plane.
1830s
W. R. Hamilton
He defines $a + i b$ as an ordered pair $(a, b)$ of real numbers with rules for adding and multiplying, removing the mystery.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 4.1, Q1–Q3	Practice 4.1 · Q1
NCERT Exercise 4.1, Q4–Q7	Practice 4.1 · Q2
NCERT Exercise 4.1, Q8–Q10	Practice 4.1 · Q3
NCERT Exercise 4.1, Q11–Q13	Practice 4.1 · Q4
NCERT Exercise 4.1, Q14	Practice 4.1 · Q5
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercises on Chapter 4 · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercises on Chapter 4 · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercises on Chapter 4 · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercises on Chapter 4 · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercises on Chapter 4 · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercises on Chapter 4 · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous exercises on Chapter 4 · Q7
NCERT Miscellaneous Exercise, Q8	Miscellaneous exercises on Chapter 4 · Q8
NCERT Miscellaneous Exercise, Q9	Miscellaneous exercises on Chapter 4 · Q9
NCERT Miscellaneous Exercise, Q10	Miscellaneous exercises on Chapter 4 · Q10
NCERT Miscellaneous Exercise, Q11	Miscellaneous exercises on Chapter 4 · Q11
NCERT Miscellaneous Exercise, Q12	Miscellaneous exercises on Chapter 4 · Q12
NCERT Miscellaneous Exercise, Q13	Miscellaneous exercises on Chapter 4 · Q13
NCERT Miscellaneous Exercise, Q14	Miscellaneous exercises on Chapter 4 · Q14

Quick check: MCQs on Complex Numbers and Quadratic Equations

4.2 Complex Numbers3 questions
4.3.4 Division of two complex numbers3 questions
4.3.7 Identities3 questions
4.4 The Modulus and the Conjugate of a Complex Number4 questions
4.5 Argand Plane and Polar Representation3 questions
Miscellaneous Examples3 questions

Glossary · 12

Key terms (English–Hindi)

additive inverseयोज्य प्रतिलोमyojya pratilom: For $z = a + i b$ , the number $- z = - a - i b$ ; the two add up to $0$ .First seen in the chapter →
Argand planeआर्गंड तलargand tal: A coordinate plane in which the point $(a, b)$ stands for the complex number $a + i b$ .First seen in the chapter →

complex numberसम्मिश्र संख्याsammishra sankhya: A number $a + i b$ in which $a$ and $b$ are real and $i^{2} = - 1$ .First seen in the chapter →
conjugateसंयुग्मीsanyugmi: For $z = a + i b$ , the number $\overset{ˉ}{z} = a - i b$ , with the sign of the imaginary part reversed.First seen in the chapter →

imaginary axisकाल्पनिक अक्षkalpanik aksh: The vertical axis of the Argand plane; its points are the numbers $0 + i b$ .First seen in the chapter →
imaginary partकाल्पनिक भागkalpanik bhag: For $z = a + i b$ , the real number $b$ (not $i b$ ), written $Im z$ .First seen in the chapter →
imaginary unitकाल्पनिक इकाईkalpanik ikai: The number $i$ whose square is $- 1$ . It is written $\sqrt{- 1}$ and read 'iota'.First seen in the chapter →

modulusमापांकmaapaank: For $z = a + i b$ , the non-negative real number $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ , the distance of $z$ from the origin.First seen in the chapter →
multiplicative inverseगुणात्मक प्रतिलोमgunatmak pratilom: For $z \neq 0$ , the number $z^{- 1}$ with $z z^{- 1} = 1$ . It equals $\overset{ˉ}{z} / ∣ z ∣^{2}$ .First seen in the chapter →

purely imaginary numberशुद्ध काल्पनिक संख्याshuddh kalpanik sankhya: A complex number $0 + i b$ with $b \neq 0$ , such as $7 i$ .First seen in the chapter →

real axisवास्तविक अक्षvastavik aksh: The horizontal axis of the Argand plane; its points are the real numbers $a + 0 i$ .First seen in the chapter →
real partवास्तविक भागvastavik bhag: For $z = a + i b$ , the real number $a$ , written $Re z$ .First seen in the chapter →

Complex Numbers and Quadratic Equations: Class 11 Maths · Chapter 4

What you will learn in Complex Numbers and Quadratic Equations

4.1 Introduction

4.2 Complex Numbers

4.3 Algebra of Complex Numbers

4.3.1 Addition of two complex numbers

4.3.2 Difference of two complex numbers

4.3.3 Multiplication of two complex numbers

4.3.4 Division of two complex numbers

4.3.5 Power of i

4.3.6 The square roots of a negative real number

4.3.7 Identities

4.4 The Modulus and the Conjugate of a Complex Number

Practice 4.1: questions with answers

4.5 Argand Plane and Polar Representation

Miscellaneous Examples

Practice questions with answers (NCERT “Miscellaneous exercises on Chapter 4”)

Complex Numbers and Quadratic Equations: key points and summary

Find your NCERT exercise question

Quick check: MCQs on Complex Numbers and Quadratic Equations

Key terms (English–Hindi)

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