The Hook: Every Number Has a GPS Coordinate!
Think about a radar screen in movies like Top Gun or Mission: Impossible. They don’t show positions as (x, y) coordinates — they show distance and angle from the radar!
Complex numbers work the same way! Instead of $z = a + ib$ (rectangular), we can use polar form: distance from origin and angle from the positive real axis.
Real-world use: GPS systems, radar, sonar, and AC circuit analysis all use polar coordinates!
Why this matters for JEE: Polar form makes multiplication, division, powers, and roots incredibly simple. Expect 2-3 questions using modulus-argument properties!
Prerequisites
Before learning modulus and argument, review:
- Complex Numbers Basics — $z = a + ib$, algebra, conjugate
- Argand Diagram — Geometric representation
- Trigonometry — $\sin, \cos, \tan$, inverse trig functions
- Pythagorean theorem — $a^2 + b^2 = c^2$
Modulus: The Distance from Origin
Definition
The modulus (or absolute value) of a complex number $z = a + ib$ is its distance from the origin on the Argand plane:
$$\boxed{|z| = \sqrt{a^2 + b^2}}$$In simple terms: Modulus is the length of the line segment from origin to the point representing $z$.
Interactive Demo: Visualize Modulus and Argument
Explore how modulus and argument define complex numbers in polar form.
Examples
| Complex Number | Modulus |
|---|---|
| $3 + 4i$ | $\sqrt{9 + 16} = 5$ |
| $-5 + 12i$ | $\sqrt{25 + 144} = 13$ |
| $1 - i$ | $\sqrt{1 + 1} = \sqrt{2}$ |
| $7$ | $\sqrt{49 + 0} = 7$ |
| $-3i$ | $\sqrt{0 + 9} = 3$ |
Memorize these for quick calculation:
- 3-4-5: $|3 + 4i| = 5$
- 5-12-13: $|5 + 12i| = 13$
- 8-15-17: $|8 + 15i| = 17$
- 7-24-25: $|7 + 24i| = 25$
These appear frequently in JEE!
Properties of Modulus
Basic Properties
Non-negative: $|z| \geq 0$, and $|z| = 0 \iff z = 0$
Conjugate: $|\bar{z}| = |z|$ (conjugate has same modulus)
Product: $z \cdot \bar{z} = |z|^2$
Proof: $(a + ib)(a - ib) = a^2 + b^2 = |z|^2$
Multiplicative: $|z_1 z_2| = |z_1| |z_2|$
Division: $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$ (if $z_2 \neq 0$)
Powers: $|z^n| = |z|^n$
Triangle Inequalities
For any complex numbers $z_1, z_2$:
$$\boxed{|z_1 + z_2| \leq |z_1| + |z_2|}$$(Triangle Inequality)
$$\boxed{|z_1 - z_2| \geq ||z_1| - |z_2||}$$ $$\boxed{|z_1 + z_2| \geq ||z_1| - |z_2||}$$Geometric meaning: In any triangle, the length of one side is less than or equal to the sum of the other two sides.
Argument: The Angle from Positive Real Axis
Definition
The argument of a non-zero complex number $z = a + ib$ is the angle $\theta$ that the line segment from origin to $z$ makes with the positive real axis.
$$\boxed{\arg(z) = \theta}$$Notation: $\arg(z)$ or sometimes $\text{amp}(z)$ (amplitude)
Finding the Argument
For $z = a + ib$:
$$\tan\theta = \frac{b}{a}$$But be careful! The formula $\theta = \tan^{-1}(b/a)$ doesn’t always give the correct angle.
The Correct Formula (Quadrant-Aware)
| Quadrant | Condition | Argument $\theta$ |
|---|---|---|
| I | $a > 0, b > 0$ | $\theta = \tan^{-1}\left(\frac{b}{a}\right)$ |
| II | $a < 0, b > 0$ | $\theta = \pi - \tan^{-1}\left(\frac{ |
| III | $a < 0, b < 0$ | $\theta = -\pi + \tan^{-1}\left(\frac{ |
| IV | $a > 0, b < 0$ | $\theta = -\tan^{-1}\left(\frac{ |
| Positive real axis | $b = 0, a > 0$ | $\theta = 0$ |
| Negative real axis | $b = 0, a < 0$ | $\theta = \pi$ or $-\pi$ |
| Positive imaginary axis | $a = 0, b > 0$ | $\theta = \frac{\pi}{2}$ |
| Negative imaginary axis | $a = 0, b < 0$ | $\theta = -\frac{\pi}{2}$ |
Wrong: $\arg(-1 + i) = \tan^{-1}(1/(-1)) = \tan^{-1}(-1) = -\pi/4$
Right: $-1 + i$ is in Quadrant II, so:
$$\theta = \pi - \tan^{-1}\left(\frac{1}{1}\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$Always plot the point to determine the correct quadrant!
Principal Argument
The principal value of the argument lies in the range:
$$\boxed{-\pi < \arg(z) \leq \pi}$$This is the standard convention in JEE.
General Argument
If $\theta$ is the principal argument, the general argument is:
$$\arg(z) = \theta + 2n\pi, \quad n \in \mathbb{Z}$$Polar Form of Complex Numbers
Conversion from Rectangular to Polar
Any complex number $z = a + ib$ can be written as:
$$\boxed{z = r(\cos\theta + i\sin\theta)}$$where:
- $r = |z| = \sqrt{a^2 + b^2}$ (modulus)
- $\theta = \arg(z)$ (argument)
This is called the polar form or trigonometric form.
Conversion from Polar to Rectangular
Given $z = r(\cos\theta + i\sin\theta)$:
$$a = r\cos\theta, \quad b = r\sin\theta$$So $z = r\cos\theta + ir\sin\theta = a + ib$
Examples
Example 1: Convert $1 + i$ to polar form.
$r = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$
$\theta = \tan^{-1}(1/1) = \pi/4$ (Quadrant I)
Answer: $z = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$
Example 2: Convert $-\sqrt{3} + i$ to polar form.
$r = \sqrt{3 + 1} = 2$
$\theta = \pi - \tan^{-1}(1/\sqrt{3}) = \pi - \pi/6 = 5\pi/6$ (Quadrant II)
Answer: $z = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)$
Exponential Form (Euler’s Formula)
Euler’s Formula
One of the most beautiful formulas in mathematics:
$$\boxed{e^{i\theta} = \cos\theta + i\sin\theta}$$This connects exponential functions, trigonometry, and complex numbers!
Exponential Form
Using Euler’s formula, the polar form becomes:
$$\boxed{z = re^{i\theta}}$$where $r = |z|$ and $\theta = \arg(z)$.
Example: $1 + i = \sqrt{2}e^{i\pi/4}$
Important Special Cases
$$e^{i\pi} = \cos\pi + i\sin\pi = -1 + 0i = -1$$This gives Euler’s Identity, considered the most beautiful equation:
$$\boxed{e^{i\pi} + 1 = 0}$$It connects five fundamental constants: $e, i, \pi, 1, 0$!
Properties of Argument
For non-zero complex numbers $z_1, z_2$:
Addition Rule
$$\boxed{\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)}$$(up to multiples of $2\pi$)
In simple terms: When you multiply complex numbers, their moduli multiply and their arguments add!
Division Rule
$$\boxed{\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)}$$Conjugate Rule
$$\boxed{\arg(\bar{z}) = -\arg(z)}$$(Reflection across real axis reverses the angle)
Negative Rule
$$\boxed{\arg(-z) = \arg(z) \pm \pi}$$Power Rule
$$\boxed{\arg(z^n) = n \cdot \arg(z)}$$(up to multiples of $2\pi$)
Multiplication and Division in Polar Form
Multiplication
If $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$:
$$\boxed{z_1 z_2 = r_1r_2[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)]}$$Rule: Multiply moduli, add arguments!
Example: Multiply $z_1 = 2(\cos 30° + i\sin 30°)$ and $z_2 = 3(\cos 45° + i\sin 45°)$
$z_1 z_2 = (2 \times 3)[\cos(30° + 45°) + i\sin(30° + 45°)]$
$= 6(\cos 75° + i\sin 75°)$
Division
$$\boxed{\frac{z_1}{z_2} = \frac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)]}$$Rule: Divide moduli, subtract arguments!
When you need to multiply or divide complex numbers:
- Convert to polar form
- Multiply/divide moduli
- Add/subtract arguments
- Convert back if needed
This is much faster than algebraic multiplication, especially for JEE Advanced!
Common Mistakes to Avoid
Wrong: $|z_1 + z_2| = |z_1| + |z_2|$
Right: $|z_1 + z_2| \leq |z_1| + |z_2|$ (Triangle Inequality)
Equality holds only when $z_1$ and $z_2$ have the same argument (same direction).
Always check which quadrant the complex number is in!
For $z = -1 - i$:
- Wrong: $\arg(z) = \tan^{-1}(1) = \pi/4$
- Right: $z$ is in Quadrant III, so $\arg(z) = -\pi + \pi/4 = -3\pi/4$
$\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$ is true up to multiples of $2\pi$.
The result might need adjustment to fit in $(-\pi, \pi]$.
Example: If $\arg(z_1) = \frac{3\pi}{4}$ and $\arg(z_2) = \frac{3\pi}{4}$:
$\arg(z_1z_2) = \frac{3\pi}{4} + \frac{3\pi}{4} = \frac{3\pi}{2}$
But principal value is $\frac{3\pi}{2} - 2\pi = -\frac{\pi}{2}$
Right: $|z_1 z_2| = |z_1| |z_2|$ (always true!)
But $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$ needs care with $2\pi$ multiples.
When to Use Polar Form
Use rectangular form $z = a + ib$ when:
- Adding or subtracting complex numbers
- Finding real and imaginary parts
- Simple algebraic manipulations
Use polar form $z = r(\cos\theta + i\sin\theta)$ when:
- Multiplying or dividing complex numbers
- Finding powers and roots
- Geometric problems involving angles
- Using De Moivre’s theorem
Key insight: Choose the form that makes your calculation easier!
Practice Problems
Level 1: Foundation (NCERT)
Find the modulus and argument of $z = 1 - i$.
Solution:
Modulus: $|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$
Argument: $z$ is in Quadrant IV ($a > 0, b < 0$)
$\theta = -\tan^{-1}\left(\frac{1}{1}\right) = -\frac{\pi}{4}$
Answer: $|z| = \sqrt{2}$, $\arg(z) = -\frac{\pi}{4}$
Express $z = 1 + \sqrt{3}i$ in polar form.
Solution:
$r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$
$\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$ (Quadrant I)
Answer: $z = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$
If $|z| = 5$ and $\arg(z) = \frac{2\pi}{3}$, find $z$ in rectangular form.
Solution:
$z = 5\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$
$= 5\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)$
$= -\frac{5}{2} + \frac{5\sqrt{3}}{2}i$
Answer: $z = -\frac{5}{2} + \frac{5\sqrt{3}}{2}i$
Level 2: JEE Main
If $z_1 = 2(\cos 60° + i\sin 60°)$ and $z_2 = 3(\cos 30° + i\sin 30°)$, find $z_1 z_2$ in polar form.
Solution:
Using multiplication rule:
$z_1 z_2 = (2)(3)[\cos(60° + 30°) + i\sin(60° + 30°)]$
$= 6(\cos 90° + i\sin 90°)$
$= 6(0 + i \cdot 1)$
$= 6i$
Answer: $6(\cos 90° + i\sin 90°)$ or $6i$
Find the argument of $\frac{1 + i}{1 - i}$.
Solution:
Method 1 (Direct):
$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i$
$\arg(i) = \frac{\pi}{2}$
Method 2 (Using argument property):
$\arg\left(\frac{1+i}{1-i}\right) = \arg(1+i) - \arg(1-i)$
$= \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}$
Answer: $\frac{\pi}{2}$
If $|z - 3| = 3$, find the maximum value of $|z|$.
Solution:
$|z - 3| = 3$ represents a circle centered at $3$ (on real axis) with radius $3$.
The point on this circle farthest from the origin is at distance:
$|z|_{\text{max}} = |3| + 3 = 6$
(This is the point $z = 6$ on the real axis)
Answer: $6$
Alternatively: Using triangle inequality: $|z| = |z - 3 + 3| \leq |z - 3| + |3| = 3 + 3 = 6$
Level 3: JEE Advanced
If $z_1, z_2$ are two complex numbers such that $|z_1| = |z_2|$ and $\arg(z_1) + \arg(z_2) = 0$, prove that $z_1 = \overline{z_2}$.
Solution:
Let $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$
Given: $r_1 = r_2 = r$ and $\theta_1 + \theta_2 = 0$
So $\theta_1 = -\theta_2$
$$z_1 = r[\cos(-\theta_2) + i\sin(-\theta_2)]$$Using: $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$
$$z_1 = r[\cos\theta_2 - i\sin\theta_2]$$But $z_2 = r(\cos\theta_2 + i\sin\theta_2)$
So $\overline{z_2} = r(\cos\theta_2 - i\sin\theta_2)$
Therefore, $z_1 = \overline{z_2}$ Proved!
If $z = \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}$, find $\arg(z)$ and express $z$ in polar form.
Solution:
Method 1 (Using argument property):
$\arg(z) = \arg(1 - i\sqrt{3}) - \arg(1 + i\sqrt{3})$
For $1 - i\sqrt{3}$: Quadrant IV, $\arg = -\tan^{-1}(\sqrt{3}) = -\frac{\pi}{3}$
For $1 + i\sqrt{3}$: Quadrant I, $\arg = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$
$\arg(z) = -\frac{\pi}{3} - \frac{\pi}{3} = -\frac{2\pi}{3}$
For modulus: $|1 - i\sqrt{3}| = \sqrt{1 + 3} = 2$
$|1 + i\sqrt{3}| = \sqrt{1 + 3} = 2$
$|z| = \frac{2}{2} = 1$
Answer: $z = \cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right) = e^{-i2\pi/3}$
$\arg(z) = -\frac{2\pi}{3}$
Prove that if $|z_1 + z_2| = |z_1| + |z_2|$, then $\arg(z_1) = \arg(z_2)$.
Solution:
The equality $|z_1 + z_2| = |z_1| + |z_2|$ in the triangle inequality holds when $z_1$ and $z_2$ are in the same direction from origin.
This means $z_2 = k z_1$ for some positive real $k > 0$.
Let $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$
If $z_2 = k z_1$, then: $r_2 e^{i\theta_2} = k r_1 e^{i\theta_1}$
Comparing moduli: $r_2 = kr_1$ ✓
Comparing arguments: $e^{i\theta_2} = e^{i\theta_1}$
Therefore: $\theta_2 = \theta_1$ (up to multiples of $2\pi$)
Hence $\arg(z_1) = \arg(z_2)$ Proved!
Quick Revision Box
| Concept | Formula | Notes |
|---|---|---|
| Modulus | $\|z\| = \sqrt{a^2 + b^2}$ | Distance from origin |
| Argument | $\tan\theta = \frac{b}{a}$ | Check quadrant! |
| Polar form | $z = r(\cos\theta + i\sin\theta)$ | $r = \|z\|, \theta = \arg(z)$ |
| Exponential form | $z = re^{i\theta}$ | Euler’s formula |
| Multiplication | $\|z_1z_2\| = \|z_1\|\|z_2\|$ | Multiply moduli |
| $\arg(z_1z_2) = \arg(z_1) + \arg(z_2)$ | Add arguments | |
| Division | $\|\frac{z_1}{z_2}\| = \frac{\|z_1\|}{\|z_2\|}$ | Divide moduli |
| $\arg(\frac{z_1}{z_2}) = \arg(z_1) - \arg(z_2)$ | Subtract arguments | |
| Conjugate | $\arg(\bar{z}) = -\arg(z)$ | Reflection |
| Triangle inequality | $\|z_1 + z_2\| \leq \|z_1\| + \|z_2\|$ | Fundamental |
Related Topics
Within Complex Numbers Chapter
- Complex Numbers Basics — Foundation
- Argand Diagram — Geometric visualization
- De Moivre’s Theorem — Powers using polar form
- Cube Roots of Unity — Special angles
Math Connections
- Trigonometry — $\sin, \cos$ and angles
- Coordinate Geometry — Distance formula
- Exponential Functions — $e^x$ and derivatives
Physics Applications
- AC Circuits — Impedance in polar form
- Wave Motion — Amplitude and phase
- Oscillations — Simple Harmonic Motion representation
Teacher’s Summary
- Modulus $|z|$ is the distance from origin; use Pythagorean theorem
- Argument $\arg(z)$ is the angle from positive real axis; always check quadrant!
- Polar form $z = r(\cos\theta + i\sin\theta)$ uses modulus and argument
- Euler’s formula $e^{i\theta} = \cos\theta + i\sin\theta$ gives exponential form
- Multiplication/Division: Multiply/divide moduli, add/subtract arguments — polar form makes this trivial!
“Polar form is the secret weapon for complex number calculations — master it and JEE problems become simple!”
Exam Strategy: For multiplication, division, powers, roots — immediately convert to polar form. For addition, subtraction — use rectangular form. Choose wisely!