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Intermediate

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Complex Numbers

Master Complex Numbers for JEE Main, JEE Advanced, NEET & Board Exams. Covers modulus, argument, polar form, Euler's formula, Argand plane, De Moivre's theorem, cube roots of unity and nth roots of unity with solved examples.

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A complex number is an expression of the form $z = a + i b$ , where $a, b \in R$ and $i = \sqrt{- 1}$ is the imaginary unit. Complex numbers extend the real number system to include solutions of equations like $x^{2} + 1 = 0$ , which have no real roots. They are indispensable in JEE (Main & Advanced), with questions spanning the Argand plane, modulus-argument form, De Moivre's theorem, and roots of unity. For NEET and board exams, the focus is on basic operations, conjugates, and modulus.

1. Basic Definitions

For a complex number $z = a + i b$ :

Real part: $Re (z) = a$
Imaginary part: $Im (z) = b$ (note: $b$ is real, not $i b$ )
Purely real: $b = 0$ , e.g., $z = 5$
Purely imaginary: $a = 0$ , e.g., $z = 3 i$
Zero complex number: $a = 0$ and $b = 0$

Powers of $i$

The powers of $i$ cycle with period 4:

Power	$i^{1}$	$i^{2}$	$i^{3}$	$i^{4}$	$i^{5}$	$i^{4 k}$
Value	$i$	$- 1$	$- i$	$1$	$i$	$1$

To find $i^{n}$ : divide $n$ by $4$ and use the remainder. If remainder is $0 \to 1$ , $1 \to i$ , $2 \to - 1$ , $3 \to - i$ .

For negative powers: $i^{- 1} = - i$ , $i^{- 2} = - 1$ , $i^{- 3} = i$ , $i^{- 4} = 1$ .

Equality of Complex Numbers

Two complex numbers $z_{1} = a + i b$ and $z_{2} = c + i d$ are equal if and only if:

$a = c$ and $b = d$

i.e., real parts are equal AND imaginary parts are equal. The concept of "greater than" or "less than" does NOT apply to complex numbers — they cannot be ordered.

2. Algebraic Operations on Complex Numbers

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

Operation	Result
Addition	$z_{1} + z_{2} = (a + c) + i (b + d)$
Subtraction	$z_{1} - z_{2} = (a - c) + i (b - d)$
Multiplication	$z_{1} z_{2} = (a c - b d) + i (a d + b c)$
Division	$\frac{z_{1}}{z_{2}} = \frac{z_{1} {\bar{z}}_{2}}{\| z_{2} \|^{2}} = \frac{(a c + b d) + i (b c - a d)}{c^{2} + d^{2}}$

Key trick for division: Multiply numerator and denominator by the conjugate of the denominator ${\bar{z}}_{2} = c - i d$ to eliminate $i$ from the denominator.

3. Conjugate of a Complex Number

The conjugate of $z = a + i b$ is $\bar{z} = a - i b$ . Geometrically, $\bar{z}$ is the reflection of $z$ in the real axis on the Argand plane.

Properties of Conjugates

Property	Result
$z + \bar{z}$	$2 a = 2 Re (z)$ (always real)
$z - \bar{z}$	$2 i b = 2 i Im (z)$ (always imaginary)
$z \cdot \bar{z}$	$a^{2} + b^{2} = \| z \|^{2}$ (always real and non-negative)
$\overset{―}{z_{1} + z_{2}}$	${\bar{z}}_{1} + {\bar{z}}_{2}$
$\overset{―}{z_{1} \cdot z_{2}}$	${\bar{z}}_{1} \cdot {\bar{z}}_{2}$
$\bar{\bar{z}}$	$z$
$z$ is real	$\Leftrightarrow z = \bar{z}$
$z$ is purely imaginary	$\Leftrightarrow z + \bar{z} = 0$

4. Modulus of a Complex Number

The modulus (or absolute value) of $z = a + i b$ is the distance from the origin to the point $(a, b)$ on the Argand plane:

$| z | = \sqrt{a^{2} + b^{2}} \geq 0$

Properties of Modulus

$| z | = 0 \Leftrightarrow z = 0$
$| z | = | \bar{z} |$
$z \bar{z} = | z |^{2}$
$| z_{1} z_{2} | = | z_{1} | | z_{2} |$
$| \frac{z_{1}}{z_{2}} | = \frac{| z_{1} |}{| z_{2} |}$ , $z_{2} \neq 0$
$| z_{1} + z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} + 2 Re (z_{1} {\bar{z}}_{2})$

Triangle Inequality

$| z_{1} + z_{2} | \leq | z_{1} | + | z_{2} |$

$| z_{1} - z_{2} | \geq | | z_{1} | - | z_{2} | |$

These are extremely useful in JEE problems involving maximum and minimum values of $| z |$ .

5. Argument of a Complex Number

The argument (or amplitude) of $z = a + i b$ is the angle $θ$ that the line $O P$ (from origin $O$ to point $P (a, b)$ ) makes with the positive real axis:

$\tan θ = \frac{b}{a}$

The principal argument $Arg (z) \in (- π, π]$ . The general argument is $\arg (z) = Arg (z) + 2 k π$ , $k \in Z$ .

Argument in Different Quadrants

Quadrant	Sign of $a, b$	Principal Argument
I	$a > 0, b > 0$	$θ = \tan^{- 1} (\frac{b}{a})$
II	$a < 0, b > 0$	$θ = π - \tan^{- 1} (\frac{b}{\| a \|})$
III	$a < 0, b < 0$	$θ = - π + \tan^{- 1} (\frac{\| b \|}{\| a \|})$
IV	$a > 0, b < 0$	$θ = - \tan^{- 1} (\frac{\| b \|}{a})$

Properties of Argument

$\arg (z_{1} z_{2}) = \arg (z_{1}) + \arg (z_{2})$
$\arg (\frac{z_{1}}{z_{2}}) = \arg (z_{1}) - \arg (z_{2})$
$\arg (\bar{z}) = - \arg (z)$
$\arg (z^{n}) = n \arg (z)$

6. Polar (Modulus-Argument) Form and Euler's Form

Every complex number can be written in polar form using its modulus $r = | z |$ and argument $θ = \arg (z)$ :

$z = r (\cos θ + i \sin θ) = r cis θ$

Using Euler's formula $e^{i θ} = \cos θ + i \sin θ$ , this becomes:

$z = r e^{i θ}$

Conversion between forms:

Cartesian to Polar: $r = \sqrt{a^{2} + b^{2}}$ , $θ = \arg (z)$
Polar to Cartesian: $a = r \cos θ$ , $b = r \sin θ$

Multiplication and Division in Polar Form

For $z_{1} = r_{1} e^{i θ_{1}}$ and $z_{2} = r_{2} e^{i θ_{2}}$ :

$z_{1} z_{2} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$

$\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} e^{i (θ_{1} - θ_{2})}$

Multiplication rotates by $θ_{2}$ and scales by $r_{2}$ . This geometric interpretation is crucial for JEE Advanced problems.

Important Special Values

$z$	$\| z \|$	$\arg (z)$	Polar Form
$1$	$1$	$0$	$e^{i \cdot 0}$
$- 1$	$1$	$π$	$e^{i π}$
$i$	$1$	$π / 2$	$e^{i π / 2}$
$- i$	$1$	$- π / 2$	$e^{- i π / 2}$
$1 + i$	$\sqrt{2}$	$π / 4$	$\sqrt{2} e^{i π / 4}$
$1 - i$	$\sqrt{2}$	$- π / 4$	$\sqrt{2} e^{- i π / 4}$

7. The Argand Plane

The Argand plane (complex plane) represents $z = a + i b$ as the point $(a, b)$ , with the $x$ -axis as the real axis and the $y$ -axis as the imaginary axis.

Geometric Representations

$| z - z_{0} | = r$ — Circle of radius $r$ centred at $z_{0}$ .
$| z - z_{1} | = | z - z_{2} |$ — Perpendicular bisector of segment joining $z_{1}$ and $z_{2}$ .
$| z - z_{1} | + | z - z_{2} | = k$ (constant $> | z_{1} - z_{2} |$ ) — Ellipse with foci at $z_{1}$ and $z_{2}$ .
$\arg (z - z_{0}) = θ$ — Ray from $z_{0}$ at angle $θ$ to the real axis.
$\arg (\frac{z - z_{1}}{z - z_{2}}) = θ$ — Arc of a circle through $z_{1}$ and $z_{2}$ on which the inscribed angle is $θ$ .

Rotation Formula

If a point $z_{1}$ is rotated about another point $z_{0}$ by angle $α$ (anticlockwise) to reach $z_{2}$ , then:

$z_{2} - z_{0} = (z_{1} - z_{0}) e^{i α}$

This is one of the most powerful tools in JEE geometry problems involving complex numbers.

Section Formula

The point dividing the segment joining $z_{1}$ and $z_{2}$ in the ratio $m : n$ :

Internally: $z = \frac{m z_{2} + n z_{1}}{m + n}$
Externally: $z = \frac{m z_{2} - n z_{1}}{m - n}$
Midpoint: $z = \frac{z_{1} + z_{2}}{2}$

8. De Moivre's Theorem

For any integer $n$ (and rational $n$ for principal values):

$(\cos θ + i \sin θ)^{n} = \cos n θ + i \sin n θ$

Equivalently: $(e^{i θ})^{n} = e^{i n θ}$ .

Valid for all integers $n$ . For rational $n = p / q$ , it gives one of the values.
Used to find powers and roots of complex numbers, and to prove trigonometric identities.

Application: Finding $n^{th}$ Roots of a Complex Number

The $n^{th}$ roots of $z = r e^{i θ}$ are:

$z_{k} = r^{1 / n} e^{i (\frac{θ + 2 k π}{n})},$ $k = 0, 1, 2, \dots, n - 1$

These $n$ roots are equally spaced on a circle of radius $r^{1 / n}$ centred at the origin, separated by angle $\frac{2 π}{n}$ .

9. Cube Roots of Unity

The three cube roots of unity (solutions of $z^{3} = 1$ ) are:

$1, ω = \frac{- 1 + i \sqrt{3}}{2}, ω^{2} = \frac{- 1 - i \sqrt{3}}{2}$

where $ω = e^{2 π i / 3}$ .

Key Properties of $ω$

$ω^{3} = 1$
$1 + ω + ω^{2} = 0$ (sum of all cube roots of unity = 0)
$ω^{2} = \bar{ω}$ (they are conjugates of each other)
$| ω | = | ω^{2} | = 1$ (both lie on the unit circle)
$ω$ and $ω^{2}$ are roots of $z^{2} + z + 1 = 0$
If $1 + ω^{n} + ω^{2 n} = {\begin{cases} 3 & if n is a multiple of 3 \\ 0 & otherwise \end{cases}$

Factorisation Using $ω$

$a^{3} - b^{3} = (a - b) (a - ω b) (a - ω^{2} b)$
$a^{3} + b^{3} = (a + b) (a + ω b) (a + ω^{2} b)$
$a^{2} + a b + b^{2} = (a - ω b) (a - ω^{2} b)$
$a^{2} + b^{2} + c^{2} - a b - b c - c a$
$= (a + ω b + ω^{2} c) (a + ω^{2} b + ω c)$

10. $n^{th}$ Roots of Unity

The $n^{th}$ roots of unity are the $n$ solutions of $z^{n} = 1$ :

$z_{k} = e^{2 π i k / n} = \cos \frac{2 k π}{n} + i \sin \frac{2 k π}{n},$
$k = 0, 1, \dots, n - 1$

Let $α = e^{2 π i / n}$ (primitive $n^{th}$ root). Then the roots are $1, α, α^{2}, \dots, α^{n - 1}$ .

Sum of all $n^{th}$ roots of unity: $1 + α + α^{2} + \dots + α^{n - 1} = 0$
Product of all $n^{th}$ roots of unity: $1 \cdot α \cdot α^{2} \dots α^{n - 1} = (- 1)^{n + 1}$
They form the vertices of a regular $n$ -gon inscribed in the unit circle $| z | = 1$ .

The Rotation Trick — Most Powerful JEE Tool in Complex Numbers

Whenever a JEE problem involves triangles, squares, or regular polygons in the Argand plane — or asks to prove that points form a particular geometric figure — use the rotation formula:

$\frac{z_{3} - z_{1}}{z_{2} - z_{1}} = \frac{| z_{3} - z_{1} |}{| z_{2} - z_{1} |} \cdot e^{i α}$

where $α$ is the angle at $z_{1}$ from $z_{2}$ to $z_{3}$ (anticlockwise). Key shortcuts:

Equilateral triangle: If $z_{1}, z_{2}, z_{3}$ form an equilateral triangle, then
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}$
Right angle at $z_{2}$ : $\frac{z_{3} - z_{2}}{z_{1} - z_{2}}$ is purely imaginary, i.e., $Re (\frac{z_{3} - z_{2}}{z_{1} - z_{2}}) = 0$ .
$z_{1}, z_{2}, z_{3}$ are collinear if $\frac{z_{3} - z_{1}}{z_{2} - z_{1}}$ is purely real.
Multiplying by $i$ rotates a complex number by $90 °$ anticlockwise; multiplying by $- i$ rotates by $90 °$ clockwise.

JEE/NEET Power Tips — Avoid These Common Mistakes

$\sqrt{- a} \cdot \sqrt{- b} \neq \sqrt{a b}$ when both $a, b > 0$ . Correct: $\sqrt{- a} \cdot \sqrt{- b} = i \sqrt{a} \cdot i \sqrt{b} = - \sqrt{a b}$ . Always convert to $i$ form first before multiplying.
Argument of $0$ is not defined. Always check $z \neq 0$ before computing $\arg (z)$ .
$| z_{1} + z_{2} |^{2} \neq | z_{1} |^{2} + | z_{2} |^{2}$ in general. The correct expansion includes a cross term: $| z_{1} + z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} + 2 Re (z_{1} {\bar{z}}_{2})$ .
Complex numbers cannot be compared. Statements like " $z_{1} > z_{2}$ " are meaningless unless both are real.
$\arg (z_{1} z_{2}) = \arg (z_{1}) + \arg (z_{2})$ holds modulo $2 π$ — the principal arguments may not add up directly. Always reduce the result to $(- π, π]$ .
$1 + ω + ω^{2} = 0$ only for $ω \neq 1$ . This property is frequently used to simplify expressions; make sure $ω$ is a non-real cube root of unity.
When finding $i^{n}$ , always reduce the exponent modulo 4. For example, $i^{103} = i^{4 \times 25 + 3} = i^{3} = - i$ .

Practice Questions (JEE / NEET / Board Level)

Q1: If $z = \frac{1 + i}{1 - i}$ , then $z^{4}$ equals:

A) -1
B) 1
C) $i$
D) $- i$

Answer: B) 1.

Explanation:

First, simplify $z$ by multiplying the numerator and the denominator by the complex conjugate of the denominator ( $1 + i$ ):
$z = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}$

Expand the expressions:
$z = \frac{(1 + i)^{2}}{1^{2} + 1^{2}} = \frac{1 + 2 i + i^{2}}{2}$

Since $i^{2} = - 1$ , this simplifies to:
$z = \frac{1 + 2 i - 1}{2} = \frac{2 i}{2} = i$

Now, calculate $z^{4}$ :
$z^{4} = i^{4} = 1$ .

Q2: The modulus and principal argument of $z = - \sqrt{3} + i$ are respectively:

A) 2 and $\frac{5 π}{6}$
B) 2 and $\frac{π}{6}$
C) $\sqrt{2}$ and $\frac{5 π}{6}$
D) 2 and $\frac{2 π}{3}$

Answer: A) 2 and $\frac{5 π}{6}$ .

Explanation:

First, calculate the modulus ( $| z |$ ):
$| z | = \sqrt{(- \sqrt{3})^{2} + 1^{2}} = \sqrt{3 + 1} = 2$

Next, determine the principal argument. Since the real part $a = - \sqrt{3} < 0$ and the imaginary part $b = 1 > 0$ , the complex number $z$ lies in the second quadrant.

Calculate the reference angle ( $α$ ):
$α = \tan^{- 1} (| \frac{1}{- \sqrt{3}} |) = \frac{π}{6}$

For the second quadrant, the principal argument ( $θ$ ) is $π - α$ :
$θ = π - \frac{π}{6} = \frac{5 π}{6}$ .

Q3: If $ω$ is a non-real cube root of unity, find the value of $(1 - ω + ω^{2})^{6}$ .

A) 32
B) 64
C) -64
D) 16

Answer: B) 64.

Explanation:

Use the standard property of cube roots of unity: $1 + ω + ω^{2} = 0$ .
From this, we can deduce that $1 + ω^{2} = - ω$ .

Substitute this into the given expression:
$1 - ω + ω^{2} = (1 + ω^{2}) - ω$
$= - ω - ω = - 2 ω$

Now raise this result to the 6th power:
$(1 - ω + ω^{2})^{6} = (- 2 ω)^{6}$
$= (- 2)^{6} \cdot ω^{6}$
$= 64 \cdot (ω^{3})^{2}$

Since $ω^{3} = 1$ , the expression evaluates to:
$64 \times 1^{2} = 64$ .

Q4: If $| z - 2 + 2 i | = 1$ , what is the minimum value of $| z |$ ?

A) $\sqrt{2} - 1$
B) $\sqrt{8} - 1$
C) $\sqrt{8} + 1$
D) 1

Answer: B) $\sqrt{8} - 1$ .

Explanation:

The equation $| z - (2 - 2 i) | = 1$ represents a circle in the Argand plane centered at $z_{0} = 2 - 2 i$ with a radius of $r = 1$ .

Find the distance from the origin to the center of the circle ( $| z_{0} |$ ):
$| z_{0} | = \sqrt{2^{2} + (- 2)^{2}} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$ .

The minimum value of $| z |$ (the shortest distance from the origin to any point on the circle) is the distance to the center minus the radius:
$| z |_{min} = | z_{0} | - r$
$| z |_{min} = 2 \sqrt{2} - 1 = \sqrt{8} - 1$ .

Q5: The value of $\sum_{k = 1}^{10} (\sin \frac{2 π k}{11} - i \cos \frac{2 π k}{11})$ is:

A) 0
B) -1
C) 1
D) $i$

Answer: D) $i$ .

Explanation:

First, rewrite the terms inside the summation by factoring out $- i$ to convert them into standard Euler form:
$\sin \frac{2 π k}{11} - i \cos \frac{2 π k}{11} = - i (\cos \frac{2 π k}{11} + i \sin \frac{2 π k}{11})$
$= - i e^{i 2 π k / 11}$

Substitute this back into the summation:
Sum $= - i \sum_{k = 1}^{10} e^{i 2 π k / 11}$

Recall that the sum of all $n$ -th roots of unity is zero. For the 11th roots of unity:
$\sum_{k = 0}^{10} e^{i 2 π k / 11} = 0$

Isolate the sum from $k = 1$ to $10$ by moving the $k = 0$ term (which equals 1) to the other side:
$\sum_{k = 1}^{10} e^{i 2 π k / 11} = - 1$

Multiply by $- i$ to find the final value:
Sum $= - i \times (- 1) = i$ .

Q6 (JEE Advanced type): If $z_{1}$ and $z_{2}$ are two complex numbers such that $| z_{1} | = | z_{2} |$ and $\arg (z_{1}) + \arg (z_{2}) = π$ , then $z_{1} + z_{2}$ is:

A) A non-zero real number
B) A purely imaginary number
C) Zero
D) Cannot be determined

Answer: B) A purely imaginary number.

Explanation:

Let the common magnitude be $| z_{1} | = | z_{2} | = r$ and let $\arg (z_{1}) = θ$ .
Given the argument condition, $\arg (z_{2}) = π - θ$ .

Express both complex numbers in polar form:
$z_{1} = r (\cos θ + i \sin θ)$
$z_{2} = r (\cos (π - θ) + i \sin (π - θ))$

Using trigonometric identities, simplify $z_{2}$ :
$z_{2} = r (- \cos θ + i \sin θ)$

Add the two complex numbers together:
$z_{1} + z_{2} = r (\cos θ + i \sin θ) + r (- \cos θ + i \sin θ)$
$z_{1} + z_{2} = r (2 i \sin θ) = (2 r \sin θ) i$

Since the real parts cancel out completely, the resulting sum is a purely imaginary number.

Q7 (JEE Advanced type): The points $z_{1} = 4 + 3 i$ , $z_{2} = - 4 + 3 i$ and $z_{3} = - 4 - 3 i$ are vertices of a rectangle. Find the fourth vertex $z_{4}$ .

A) $4 - 3 i$
B) $4 + 3 i$
C) $- 4 + 3 i$
D) $3 + 4 i$

Answer: A) $4 - 3 i$ .

Explanation:

In a rectangle, the diagonals bisect each other, meaning the midpoint of one diagonal is the same as the midpoint of the other.

The diagonals connect opposite vertices: $z_{1}$ to $z_{3}$ and $z_{2}$ to $z_{4}$ .

Calculate the midpoint of the diagonal $z_{1} z_{3}$ :
Midpoint $= \frac{z_{1} + z_{3}}{2} = \frac{(4 + 3 i) + (- 4 - 3 i)}{2} = \frac{0}{2} = 0$

Since the midpoint of $z_{2} z_{4}$ must also be at the origin ( $0$ ):
$\frac{z_{2} + z_{4}}{2} = 0$
$z_{4} = - z_{2}$

Calculate the coordinates of the fourth vertex:
$z_{4} = - (- 4 + 3 i) = 4 - 3 i$ .

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Complex Numbers

1. Basic Definitions

Powers of i

Equality of Complex Numbers

2. Algebraic Operations on Complex Numbers

3. Conjugate of a Complex Number

Properties of Conjugates

4. Modulus of a Complex Number

Properties of Modulus

Triangle Inequality

5. Argument of a Complex Number

Argument in Different Quadrants

Properties of Argument

6. Polar (Modulus-Argument) Form and Euler's Form

Multiplication and Division in Polar Form

Important Special Values

7. The Argand Plane

Geometric Representations

Rotation Formula

Section Formula

8. De Moivre's Theorem

Application: Finding nth Roots of a Complex Number

9. Cube Roots of Unity

Key Properties of ω

Factorisation Using ω

10. nth Roots of Unity

The Rotation Trick — Most Powerful JEE Tool in Complex Numbers

Practice Questions (JEE / NEET / Board Level)

Related Study Material

Practice Questions

Topic Information

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Study Tips

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Powers of $i$

Application: Finding $n^{th}$ Roots of a Complex Number

Key Properties of $ω$

Factorisation Using $ω$

10. $n^{th}$ Roots of Unity