Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)**
Published: October 2025
Category: Complex Numbers | Geometry & Algebra
🔹 Introduction
Every complex number can be represented algebraically as z=x+iyz = x + iyz=x+iy.
But there’s a more powerful way to express it — using length and angle instead of x and y.
This is called the Modulus–Argument (Polar) Form, or Modulus–Augmented Form, of a complex number.
It connects algebra with geometry, showing complex numbers as rotations and scalings on the Argand plane.
🧭 1️⃣ Cartesian Form vs Polar Form
| Representation | Expression | Key Idea |
|---|---|---|
| Cartesian (Rectangular) | z=x+iyz = x + iyz=x+iy | Real + Imaginary parts |
| Modulus–Argument (Polar) | z=r(cosθ+isinθ)z = r(\cos\theta + i\sin\theta)z=r(cosθ+isinθ) | Length + Angle representation |
🔹 Where do r and θ come from?
For z=x+iyz = x + iyz=x+iy: r=∣z∣=x2+y2r = |z| = \sqrt{x^2 + y^2}r=∣z∣=x2+y2 θ=arg(z)=tan−1 (yx)\theta = \arg(z) = \tan^{-1}\!\left(\frac{y}{x}\right)θ=arg(z)=tan−1(xy)
✅ So we can write: z=r(cosθ+isinθ)\boxed{z = r(\cos\theta + i\sin\theta)}z=r(cosθ+isinθ)
🔹 Example 1
z=1+i3z = 1 + i\sqrt{3}z=1+i3 r=12+(3)2=2,θ=tan−1 (31)=60°=π3r = \sqrt{1^2 + (\sqrt{3})^2} = 2, \quad \theta = \tan^{-1}\!\left(\frac{\sqrt{3}}{1}\right) = 60° = \frac{\pi}{3}r=12+(3)2=2,θ=tan−1(13)=60°=3π z=2(cosπ3+isinπ3)\boxed{z = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})}z=2(cos3π+isin3π)
⚡️ 2️⃣ The Euler Form
Using Euler’s identity: eiθ=cosθ+isinθe^{i\theta} = \cos\theta + i\sin\thetaeiθ=cosθ+isinθ
we can write: z=reiθ\boxed{z = re^{i\theta}}z=reiθ
This is the exponential form of a complex number.
✅ It’s compact, elegant, and makes powers & roots easy (via De Moivre’s theorem).
🔹 Example 2
z=4(cos60°+isin60°)=4eiπ/3z = 4(\cos60° + i\sin60°) = 4e^{i\pi/3}z=4(cos60°+isin60°)=4eiπ/3
🎯 3️⃣ Conversion Between Forms
From Cartesian → Polar
If z=x+iyz = x + iyz=x+iy: r=x2+y2,θ=tan−1 (yx)r = \sqrt{x^2 + y^2}, \quad \theta = \tan^{-1}\!\left(\frac{y}{x}\right)r=x2+y2,θ=tan−1(xy)
Then z=r(cosθ+isinθ)z = r(\cos\theta + i\sin\theta)z=r(cosθ+isinθ).
From Polar → Cartesian
If z=r(cosθ+isinθ)z = r(\cos\theta + i\sin\theta)z=r(cosθ+isinθ): x=rcosθ,y=rsinθx = r\cos\theta, \quad y = r\sin\thetax=rcosθ,y=rsinθ
So z=x+iyz = x + iyz=x+iy.
🔹 Example 3
Convert z=5(cos120°+isin120°)z = 5(\cos120° + i\sin120°)z=5(cos120°+isin120°) to Cartesian form. x=5cos120°=5(−12)=−2.5x = 5\cos120° = 5(-\tfrac{1}{2}) = -2.5x=5cos120°=5(−21)=−2.5 y=5sin120°=5(32)=532y = 5\sin120° = 5(\tfrac{\sqrt{3}}{2}) = \tfrac{5\sqrt{3}}{2}y=5sin120°=5(23)=253 z=−2.5+i532\boxed{z = -2.5 + i\frac{5\sqrt{3}}{2}}z=−2.5+i253
🌀 4️⃣ Geometric Meaning
(Illustration: Argand plane with point P(r,θ)P(r, \theta)P(r,θ), vector OP = r, angle θ with real axis.)
- rrr: distance from origin (modulus)
- θ\thetaθ: angle with positive real axis (argument)
Each complex number corresponds to a point P(r,θ)P(r, \theta)P(r,θ) or vector OPOPOP from origin.
🔺 5️⃣ Principal Argument
Since angles can differ by multiples of 2π2\pi2π: General argument: θ+2nπ,n∈Z\text{General argument: } \theta + 2n\pi, \quad n \in \mathbb{Z}General argument: θ+2nπ,n∈Z
But we usually define the principal argument as: −π<arg(z)≤π\boxed{-\pi < \arg(z) \leq \pi}−π<arg(z)≤π
✅ Keeps one unique angle for each complex number.
🔹 Example 4
z=−1−iz = -1 – iz=−1−i r=2, tan−1 (−1−1)=45° but point in 3rd quadrant → θ=−135°=−3π4r = \sqrt{2}, \ \tan^{-1}\!\left(\frac{-1}{-1}\right) = 45° \text{ but point in 3rd quadrant → } \theta = -135° = -\frac{3\pi}{4}r=2, tan−1(−1−1)=45° but point in 3rd quadrant → θ=−135°=−43π z=2(cos(−3π4)+isin(−3π4))\boxed{z = \sqrt{2}(\cos(-\frac{3\pi}{4}) + i\sin(-\frac{3\pi}{4}))}z=2(cos(−43π)+isin(−43π))
🧮 6️⃣ Multiplication & Division in Polar Form
Using: z1=r1(cosθ1+isinθ1),z2=r2(cosθ2+isinθ2)z_1 = r_1(\cos\theta_1 + i\sin\theta_1), \quad z_2 = r_2(\cos\theta_2 + i\sin\theta_2)z1=r1(cosθ1+isinθ1),z2=r2(cosθ2+isinθ2)
Multiplication:
z1z2=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]z_1z_2 = r_1r_2[\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)]z1z2=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]
✅ Multiply moduli, add arguments.
Division:
z1z2=r1r2[cos(θ1−θ2)+isin(θ1−θ2)]\frac{z_1}{z_2} = \frac{r_1}{r_2}[\cos(\theta_1-\theta_2) + i\sin(\theta_1-\theta_2)]z2z1=r2r1[cos(θ1−θ2)+isin(θ1−θ2)]
✅ Divide moduli, subtract arguments.
🔹 Example 5
Let z1=2(cos45°+isin45°)z_1 = 2(\cos45° + i\sin45°)z1=2(cos45°+isin45°), z2=3(cos30°+isin30°)z_2 = 3(\cos30° + i\sin30°)z2=3(cos30°+isin30°) z1z2=6(cos75°+isin75°)z_1z_2 = 6(\cos75° + i\sin75°)z1z2=6(cos75°+isin75°) z1z2=23(cos15°+isin15°)\frac{z_1}{z_2} = \frac{2}{3}(\cos15° + i\sin15°)z2z1=32(cos15°+isin15°)
✅ Geometrically: Rotation by 45° + 30° = 75°, scaling ×3.
🧩 7️⃣ Why Polar Form Matters
| Advantage | Explanation |
|---|---|
| Simplifies powers & roots | via De Moivre’s theorem |
| Geometric interpretation | rotation + scaling |
| Compact exponential form | z=reiθz = re^{i\theta}z=reiθ |
| Foundation for loci & transformations | used in IB HL & A Level vectors |
| Links trigonometry and complex algebra | bridges geometry & analysis |
🔹 Common Mistakes
- ❌ Using degree angles in radians or vice versa.
- ❌ Forgetting sign of argument (quadrant errors).
- ❌ Ignoring principal argument range.
- ❌ Confusing modulus with argument.
🌟 Why It Matters
The modulus–argument form makes complex numbers geometrically meaningful:
they become rotations, scalings, and transformations in the plane.
It’s essential for understanding:
- De Moivre’s theorem
- Roots of unity
- Loci and transformations
- Argand diagram interpretations
Algebra + Geometry = Complex Beauty.
📘 Learn Beyond the Formula
At Math By Rishabh, complex numbers aren’t just equations — they’re pictures of algebraic beauty.
In the Mathematics Elevate Mentorship Program, you’ll:
✅ Master conversions between forms,
✅ Visualize multiplication as rotation,
✅ Derive and apply polar forms to STEP-level problems.
🚀 Think geometrically, solve algebraically.
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