The Hook: Numbers Get Coordinates!
Think about Google Maps — every location has coordinates (latitude, longitude). What if numbers could have coordinates too?
In 1806, Jean-Robert Argand had this brilliant idea: represent complex numbers as points on a plane! Just like in Oppenheimer (2023) where physicists visualized invisible atoms, we can visualize “imaginary” numbers.
Real-world application: Engineers use Argand diagrams to analyze AC circuits, signal processing, and even airplane navigation systems!
Why this matters for JEE: Argand diagram questions test your geometry skills. Expect 1-2 questions combining complex numbers with coordinate geometry!
Prerequisites
Before learning Argand diagrams, make sure you understand:
- Complex Numbers Basics — $z = a + ib$, algebra of complex numbers
- Coordinate Geometry Basics — Plotting points, distance formula
- Cartesian plane — x-axis, y-axis, quadrants
Interactive Demo: Explore the Argand Plane
Try plotting different complex numbers and see how they appear on the Argand plane! Use the sliders to adjust the real and imaginary parts, or switch to polar form to control modulus and argument. Notice how:
- The conjugate reflects across the real axis
- The negative rotates the point by 180 degrees
- The reciprocal inverts the modulus and negates the argument
Explore the presets to see common complex numbers and roots of unity!
The Argand Plane: A Map for Complex Numbers
The Big Idea
Every complex number $z = a + ib$ can be represented as a point $(a, b)$ on a 2D plane.
$$\boxed{z = a + ib \longleftrightarrow \text{Point } (a, b) \text{ on the Argand plane}}$$The Axes
- Real axis (horizontal): Represents the real part $a$
- Imaginary axis (vertical): Represents the imaginary part $b$
Imaginary Axis (Im)
↑
|
z = 3+2i • (3, 2)
|
←---------+--------→ Real Axis (Re)
|
|
In simple terms: The Argand diagram is just like a coordinate plane, but the x-axis is for real parts and y-axis is for imaginary parts!
Examples
| Complex Number | Point on Argand Plane | Location |
|---|---|---|
| $3 + 2i$ | $(3, 2)$ | Quadrant I |
| $-2 + 4i$ | $(-2, 4)$ | Quadrant II |
| $-1 - 3i$ | $(-1, -3)$ | Quadrant III |
| $5 - 2i$ | $(5, -2)$ | Quadrant IV |
| $4$ | $(4, 0)$ | On real axis |
| $3i$ | $(0, 3)$ | On imaginary axis |
| $0$ | $(0, 0)$ | Origin |
Special Subsets on Argand Plane
Real Numbers
All real numbers lie on the real axis (imaginary part = 0).
Points: $(a, 0)$ for $a \in \mathbb{R}$
Pure Imaginary Numbers
All pure imaginary numbers lie on the imaginary axis (real part = 0).
Points: $(0, b)$ for $b \in \mathbb{R}, b \neq 0$
Origin
The complex number $0 = 0 + 0i$ is at the origin $(0, 0)$.
Geometric Interpretation of Operations
Addition: Vector Addition
Adding two complex numbers $z_1 = a_1 + ib_1$ and $z_2 = a_2 + ib_2$ is like vector addition:
$$z_1 + z_2 = (a_1 + a_2) + i(b_1 + b_2)$$Geometrically, complete the parallelogram with $z_1$ and $z_2$ as adjacent sides. The diagonal gives $z_1 + z_2$!
If $O$ is origin, and $A, B$ represent $z_1, z_2$, then the point $C$ representing $z_1 + z_2$ completes the parallelogram $OACB$.
This is the parallelogram law of vector addition!
Subtraction: Vector Subtraction
$$z_1 - z_2 = (a_1 - a_2) + i(b_1 - b_2)$$Geometrically, $z_1 - z_2$ is the vector from point $z_2$ to point $z_1$.
Conjugate: Reflection Across Real Axis
The conjugate $\bar{z} = a - ib$ is the mirror image of $z = a + ib$ across the real axis.
If $z$ is at $(a, b)$, then $\bar{z}$ is at $(a, -b)$.
Negative: Reflection Through Origin
$-z = -a - ib$ is the reflection of $z = a + ib$ through the origin.
If $z$ is at $(a, b)$, then $-z$ is at $(-a, -b)$.
This is a $180°$ rotation!
Distance on the Argand Plane
Distance from Origin: Modulus
The distance from the origin to the point $z = a + ib$ is called the modulus of $z$:
$$\boxed{|z| = \sqrt{a^2 + b^2}}$$This is just the distance formula from coordinate geometry!
Example: $|3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
The modulus is the hypotenuse of a right triangle with sides $a$ and $b$.
Think: “3-4-5 triangle” — appears frequently in JEE!
Distance Between Two Complex Numbers
The distance between $z_1 = a_1 + ib_1$ and $z_2 = a_2 + ib_2$ is:
$$\boxed{|z_1 - z_2| = \sqrt{(a_1 - a_2)^2 + (b_1 - b_2)^2}}$$In simple terms: $|z_1 - z_2|$ is the distance between points $z_1$ and $z_2$ on the Argand plane.
Example: Distance between $3 + 4i$ and $1 + i$:
$|{(3+4i) - (1+i)}| = |2 + 3i| = \sqrt{4 + 9} = \sqrt{13}$
Geometric Loci on Argand Plane
Many equations involving complex numbers represent geometric shapes!
Circle: $|z - z_0| = r$
Equation: $|z - z_0| = r$
Meaning: Set of all points at distance $r$ from the center $z_0$.
Geometric shape: Circle with center $z_0$ and radius $r$.
Example: $|z - (2 + 3i)| = 5$
This is a circle centered at $(2, 3)$ with radius $5$.
Perpendicular Bisector: $|z - z_1| = |z - z_2|$
Equation: $|z - z_1| = |z - z_2|$
Meaning: Set of all points equidistant from $z_1$ and $z_2$.
Geometric shape: Perpendicular bisector of the line segment joining $z_1$ and $z_2$.
Example: $|z - 1| = |z + 1|$ (i.e., $|z - 1| = |z - (-1)|$)
Let $z = x + iy$:
$\sqrt{(x-1)^2 + y^2} = \sqrt{(x+1)^2 + y^2}$
Squaring: $(x-1)^2 = (x+1)^2$
$x^2 - 2x + 1 = x^2 + 2x + 1$
$x = 0$
This is the imaginary axis!
Interior of Circle: $|z - z_0| < r$
Equation: $|z - z_0| < r$
Geometric shape: Interior (inside) of circle centered at $z_0$ with radius $r$.
Exterior of Circle: $|z - z_0| > r$
Equation: $|z - z_0| > r$
Geometric shape: Exterior (outside) of circle centered at $z_0$ with radius $r$.
Triangle Inequality
For any two complex numbers $z_1$ and $z_2$:
$$\boxed{|z_1 + z_2| \leq |z_1| + |z_2|}$$Geometric meaning: In a triangle, the length of any side is less than or equal to the sum of the other two sides.
Equality holds when $z_1$ and $z_2$ are in the same direction (positive real multiples).
Related Inequalities
$$|z_1 - z_2| \geq ||z_1| - |z_2||$$ $$|z_1 + z_2| \geq ||z_1| - |z_2||$$These come from the triangle inequality!
Common Mistakes to Avoid
Wrong: Thinking of $z = 3 + 4i$ as only a point
Right: $z = 3 + 4i$ can be thought of as:
- A point $(3, 4)$ on the plane
- A vector from origin to $(3, 4)$
Both interpretations are valid and useful!
Remember:
- $|z|$ = distance from origin to $z$
- $|z_1 - z_2|$ = distance from $z_1$ to $z_2$
Common error: Writing $|z_1 - z_2|$ as $|z_1| - |z_2|$ ✗
These are NOT the same!
Wrong: $|z| = 3$ is the equation of a circle with radius 3 centered at origin ✓
But: $|z - 2i| = 3$ is centered at $(0, 2)$, NOT at $(2, 0)$!
Always identify the center correctly: $z_0 = 0 + 2i = (0, 2)$
The conjugate $\bar{z}$ is reflection across the real axis (horizontal), not the imaginary axis!
If $z = 2 + 3i$, then $\bar{z} = 2 - 3i$ (same x-coordinate, opposite y-coordinate)
When to Use Argand Diagram
Use Argand diagram when:
- The problem involves geometric interpretation ($|z - z_0| = r$, etc.)
- Finding locus of points satisfying conditions
- Visualizing addition/subtraction geometrically
- Problems involving distance, circles, or perpendicular bisectors
- When algebraic approach seems too complicated!
Key insight: Convert complex number equations to coordinate geometry — it’s often easier!
Practice Problems
Level 1: Foundation (NCERT)
Plot the following complex numbers on the Argand plane: (a) $2 + 3i$ (b) $-1 + 4i$ (c) $3 - 2i$ (d) $-3 - i$
Solution:
(a) $(2, 3)$ — Quadrant I (b) $(-1, 4)$ — Quadrant II (c) $(3, -2)$ — Quadrant IV (d) $(-3, -1)$ — Quadrant III
(Plot these on graph paper for practice!)
Find the distance between $z_1 = 3 + 4i$ and $z_2 = 6 + 8i$.
Solution:
$|z_1 - z_2| = |(3+4i) - (6+8i)| = |-3 - 4i| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5$
Answer: $5$ units
What geometric shape does $|z - 2i| = 3$ represent?
Solution:
This is the equation $|z - z_0| = r$ where $z_0 = 0 + 2i = 2i$ and $r = 3$.
Answer: Circle with center $(0, 2)$ and radius $3$
Level 2: JEE Main
If $|z - 3| = |z - 5i|$, find the locus of $z$ in the Argand plane.
Solution:
This represents points equidistant from $3$ and $5i$.
$3$ is at $(3, 0)$ and $5i$ is at $(0, 5)$.
The locus is the perpendicular bisector of the line segment joining these points.
Let $z = x + iy$:
$\sqrt{(x-3)^2 + y^2} = \sqrt{x^2 + (y-5)^2}$
Squaring both sides: $(x-3)^2 + y^2 = x^2 + (y-5)^2$
$x^2 - 6x + 9 + y^2 = x^2 + y^2 - 10y + 25$
$-6x + 9 = -10y + 25$
$-6x + 10y = 16$
$3x - 5y = -8$
Answer: The line $3x - 5y + 8 = 0$
Find the area of the triangle formed by the complex numbers $0, 3 + 4i,$ and $3 - 4i$ on the Argand plane.
Solution:
Vertices are: $O(0, 0)$, $A(3, 4)$, $B(3, -4)$
Using area formula for a triangle:
$$\text{Area} = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ $$= \frac{1}{2}|0(4 - (-4)) + 3((-4) - 0) + 3(0 - 4)|$$ $$= \frac{1}{2}|0 - 12 - 12| = \frac{1}{2} \times 24 = 12$$Alternative: Base $= |AB| = 8$ (from $(3,4)$ to $(3,-4)$), Height $= 3$ (distance from origin to line $x=3$)
Area $= \frac{1}{2} \times 8 \times 3 = 12$
Answer: $12$ square units
If $|z| = 4$ and $\text{arg}(z) = \frac{\pi}{6}$, find the complex number $z$.
Solution:
From polar form: $z = r(\cos\theta + i\sin\theta)$
$z = 4\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$
$= 4\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right)$
$= 2\sqrt{3} + 2i$
Answer: $z = 2\sqrt{3} + 2i$
Verification: $|z| = \sqrt{(2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4$ ✓
Level 3: JEE Advanced
If $z_1, z_2, z_3$ are complex numbers such that $|z_1| = |z_2| = |z_3| = 1$ and $z_1 + z_2 + z_3 = 0$, find the geometric interpretation and the value of $|z_1 - z_2|^2 + |z_2 - z_3|^2 + |z_3 - z_1|^2$.
Solution:
All three lie on the unit circle: $|z_1| = |z_2| = |z_3| = 1$
Given: $z_1 + z_2 + z_3 = 0$, which means the centroid is at the origin.
For points on a unit circle with centroid at origin, they form an equilateral triangle.
Now calculate:
$$|z_1 - z_2|^2 = (z_1 - z_2)\overline{(z_1 - z_2)} = (z_1 - z_2)(\bar{z_1} - \bar{z_2})$$ $$= z_1\bar{z_1} - z_1\bar{z_2} - z_2\bar{z_1} + z_2\bar{z_2}$$ $$= |z_1|^2 + |z_2|^2 - (z_1\bar{z_2} + \bar{z_1}z_2)$$ $$= 1 + 1 - 2\text{Re}(z_1\bar{z_2}) = 2 - 2\text{Re}(z_1\bar{z_2})$$Similarly for others. Sum:
$$\sum |z_i - z_j|^2 = 6 - 2[\text{Re}(z_1\bar{z_2}) + \text{Re}(z_2\bar{z_3}) + \text{Re}(z_3\bar{z_1})]$$From $z_1 + z_2 + z_3 = 0$, take conjugate: $\bar{z_1} + \bar{z_2} + \bar{z_3} = 0$
Multiply: $(z_1 + z_2 + z_3)(\bar{z_1} + \bar{z_2} + \bar{z_3}) = 0$
Expanding and using $|z_i|^2 = z_i\bar{z_i} = 1$:
$3 + 2[\text{Re}(z_1\bar{z_2}) + \text{Re}(z_2\bar{z_3}) + \text{Re}(z_3\bar{z_1})] = 0$
So the sum in brackets $= -\frac{3}{2}$
Therefore: $\sum |z_i - z_j|^2 = 6 - 2(-\frac{3}{2}) = 6 + 3 = 9$
Answer: $9$
If $z$ satisfies $|z - 1| + |z + 1| = 4$, find the locus of $z$.
Solution:
Let $z = x + iy$. Then:
$|z - 1| = \sqrt{(x-1)^2 + y^2}$ (distance from $(1, 0)$)
$|z + 1| = \sqrt{(x+1)^2 + y^2}$ (distance from $(-1, 0)$)
The equation states: sum of distances from $(1, 0)$ and $(-1, 0)$ is $4$.
This is the definition of an ellipse with foci at $F_1(-1, 0)$ and $F_2(1, 0)$.
For an ellipse: $2a = 4 \Rightarrow a = 2$
Distance between foci: $2c = 2 \Rightarrow c = 1$
$b^2 = a^2 - c^2 = 4 - 1 = 3$
Answer: Ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$
Quick Revision Box
| Concept | Geometric Interpretation |
|---|---|
| $z = a + ib$ | Point $(a, b)$ on Argand plane |
| Real axis | All points $(a, 0)$ |
| Imaginary axis | All points $(0, b)$ |
| $\|z\|$ | Distance from origin to $z$ |
| $\|z_1 - z_2\|$ | Distance between $z_1$ and $z_2$ |
| $\bar{z}$ | Reflection of $z$ across real axis |
| $-z$ | Reflection of $z$ through origin |
| $\|z - z_0\| = r$ | Circle with center $z_0$, radius $r$ |
| $\|z - z_1\| = \|z - z_2\|$ | Perpendicular bisector of segment joining $z_1, z_2$ |
| Triangle inequality | $\|z_1 + z_2\| \leq \|z_1\| + \|z_2\|$ |
Related Topics
Within Complex Numbers Chapter
- Complex Numbers Basics — Foundation for Argand diagram
- Modulus and Argument — Polar representation
- De Moivre’s Theorem — Rotations on Argand plane
Math Connections
- Coordinate Geometry — Distance, locus, circles
- Trigonometry — Angles and argument
- Vector Algebra — Complex numbers as 2D vectors
Physics Applications
- AC Circuits — Phasor diagrams are Argand diagrams
- Wave Interference — Complex amplitude representation
- Signal Processing — Frequency domain representation
Teacher’s Summary
- Argand plane maps every complex number $z = a + ib$ to point $(a, b)$
- Modulus $|z|$ is the distance from origin; $|z_1 - z_2|$ is distance between points
- Conjugate is reflection across real axis; negative is rotation by $180°$
- Geometric loci: $|z - z_0| = r$ gives circles, $|z - z_1| = |z - z_2|$ gives perpendicular bisectors
- Triangle inequality connects algebra and geometry: $|z_1 + z_2| \leq |z_1| + |z_2|$
“The Argand diagram transforms algebra into geometry — when stuck on a complex number problem, draw it!”
Exam Strategy: Always sketch an Argand diagram for locus problems. Visual solutions are faster and less error-prone!