JEE Mains Maths · Complex Numbers

Algebra of Complex Numbers

Working with z = x + iy directly: splitting a quotient into real and imaginary parts, solving equations in z and its conjugate, and using the rules for the modulus.

All Complex Numbers notes JEE Mains Maths strategy

Why this matters

Twenty-eight PYQs, twenty-two of them multiple choice. Seven ask for a real or imaginary part, or when a quotient is purely real or purely imaginary; fourteen solve an equation in z and its conjugate; seven use the modulus rules or bound the size of z. Three ideas cover the page.

Concept 1 of 3: Real and imaginary parts

To split a quotient, multiply top and bottom by the conjugate of the denominator: the denominator becomes the real number

c^2+d^2

. Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal, so one complex equation gives two real ones. 'Purely real' sets the imaginary part to 0; 'purely imaginary' sets the real part to 0.

Definition

$z=x+iy$ : $\mathrm{Re}(z)=x$ , $\mathrm{Im}(z)=y$ ; conjugate $\bar z=x-iy$ .
$\frac{a+ib}{c+id}=\frac{(a+ib)(c-id)}{c^2+d^2}$ .
$a+ib=c+id$ exactly when $a=c$ and $b=d$ .
$i^2=-1$ , and $i^n$ depends on $n$ mod 4.

Dividing complex numbers

\frac{a+ib}{c+id}=\frac{(ac+bd)+i(bc-ad)}{c^2+d^2}

Worked example

Find real

x

and

y

with

(x+iy)(2-i)=3+4i

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 8 April 2024 · Q156Moderate

Example 1 · Complex Numbers · Algebra of Complex Numbers

The sum of all possible values of

\theta \in \lbrack - \pi,2\pi\rbrack

, for which

\frac{1 + icos\theta}{1 - 2icos\theta}

is purely imaginary, is equal to

Count every angle in the interval

A condition such as

\cos^2\theta=\frac12

has several solutions in an interval like

[-\pi,2\pi]

. List them all before adding, and check that no denominator vanishes at any of them.

Drill 6 more on real and imaginary parts

Concept 2 of 3: Equations in z and its conjugate

Put

z=x+iy

and compare real and imaginary parts: two real equations in

x

and

y

. Terms like

|z|

and

z\bar z

are real, so they sit wholly in the real part — often the imaginary part alone fixes

y

at once. For an equation such as

z^2=k\bar z

, taking the modulus of both sides first gives

|z|

directly.

Definition

$z\bar z=|z|^2$ , $z+\bar z=2\,\mathrm{Re}(z)$ , $z-\bar z=2i\,\mathrm{Im}(z)$ .
Equate real parts and imaginary parts.
Taking moduli: $|z^2|=|z|^2$ , $|\bar z|=|z|$ .
Keep $z=0$ if the equation allows it, and drop it if the question says non-zero.

Conjugate identities

z\bar z=|z|^2,\quad z+\bar z=2\,\mathrm{Re}(z),\quad z-\bar z=2i\,\mathrm{Im}(z)

Worked example

Solve

|z|+z=2+i

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q54Moderate

Example 2 · Complex Numbers · Algebra of Complex Numbers

Let

S =\left\{ z\in C:4z^{2}+\overline{z}= 0 \right\}

. Then

\sum_{z \in S} |z|^{2}

is equal to :

Squaring can add a root

\sqrt{x^2+y^2}=x+k

, squaring needs

x+k\ge0

. Check each answer in the original equation, since a modulus is never negative.

Drill 13 more on equations in z and its conjugate

Concept 3 of 3: Modulus rules and bounds

The modulus multiplies and divides:

|z_1z_2|=|z_1||z_2|

. To handle a sum, expand

|a+b|^2=(a+b)(\bar a+\bar b)

. The triangle inequality bounds a modulus from both sides, with equality when the two numbers point the same way (or opposite ways). When only

|z|

appears, call it

r

and solve a real equation or inequality in

r\ge0

Definition

$|z_1z_2|=|z_1||z_2|$ , $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$ , $|\bar z|=|z|$ .
$|a\pm b|^2=|a|^2+|b|^2\pm2\,\mathrm{Re}(a\bar b)$ .
$|a+b|^2+|a-b|^2=2(|a|^2+|b|^2)$ .
$\big||a|-|b|\big|\le|a\pm b|\le|a|+|b|$ .

Modulus of a sum

|a\pm b|^2=|a|^2+|b|^2\pm2\,\mathrm{Re}(a\bar b)

Worked example

|z_1|=3

|z_2|=4

and

|z_1+z_2|=5

, find

|z_1-z_2|

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 July 2022 · Q151Moderate

Example 3 · Complex Numbers · Algebra of Complex Numbers

z \neq 0

be a complex number such that

\left| z -\frac{1}{z} \right|= 2

, then the maximum value of

|z|

is:

A bound must be reached

The triangle inequality gives a bound, and it is the answer only if some

z

attains it. Check that the extreme case points the numbers the same way and still satisfies every condition.

Drill 6 more on modulus rules and bounds

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Real and imaginary parts
Dividing complex numbers
$\frac{a+ib}{c+id}=\frac{(ac+bd)+i(bc-ad)}{c^2+d^2}$
Equations in z and its conjugate
Conjugate identities
$z\bar z=|z|^2,\quad z+\bar z=2\,\mathrm{Re}(z),\quad z-\bar z=2i\,\mathrm{Im}(z)$
Modulus rules and bounds
Modulus of a sum
$|a\pm b|^2=|a|^2+|b|^2\pm2\,\mathrm{Re}(a\bar b)$

Watch out for (3)

Count every angle in the interval
→ Real and imaginary parts
Squaring can add a root
→ Equations in z and its conjugate
A bound must be reached
→ Modulus rules and bounds

Test yourself on Complex Numbers

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Complex Numbers chapter test Drill all 28 questions