Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)**
Published: October 2025
Category: Complex Numbers | Algebra & Geometry
🔹 Introduction
Complex numbers connect algebra and geometry like nothing else in mathematics.
Every complex number can be represented as a point or a vector on the Argand plane, and taking roots of complex numbers reveals beautiful symmetric patterns — perfect polygons centered at the origin.
Finding roots of a complex number means finding all complex numbers that, when raised to a power, give the original number.
⚡️ 1️⃣ Polar Form of a Complex Number
Every complex number can be expressed as: z=r(cosθ+isinθ)z = r(\cos\theta + i\sin\theta)z=r(cosθ+isinθ)
or, using Euler’s form, z=reiθz = re^{i\theta}z=reiθ
where
- r=∣z∣=x2+y2r = |z| = \sqrt{x^2 + y^2}r=∣z∣=x2+y2 (modulus)
- θ=arg(z)=tan−1(yx)\theta = \arg(z) = \tan^{-1}\left(\frac{y}{x}\right)θ=arg(z)=tan−1(xy) (argument)
🔹 Example
If z=1+i3z = 1 + i\sqrt{3}z=1+i3, r=12+(3)2=2,θ=tan−1(31)=π3r = \sqrt{1^2 + (\sqrt{3})^2} = 2, \quad \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}r=12+(3)2=2,θ=tan−1(13)=3π
So, z=2(cosπ3+isinπ3)z = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})z=2(cos3π+isin3π)
🧭 2️⃣ De Moivre’s Theorem
For any complex number z=r(cosθ+isinθ)z = r(\cos\theta + i\sin\theta)z=r(cosθ+isinθ) and integer nnn: zn=rn(cosnθ+isinnθ)z^n = r^n (\cos n\theta + i\sin n\theta)zn=rn(cosnθ+isinnθ)
✅ This allows us to find powers and roots easily in polar form.
🎯 3️⃣ The nnnth Roots of a Complex Number
If z=r(cosθ+isinθ)z = r(\cos\theta + i\sin\theta)z=r(cosθ+isinθ),
then all the nnnth roots of z are given by: zk=r1/n[cos(θ+2kπn)+isin(θ+2kπn)],k=0,1,2,…,n−1\boxed{z_k = r^{1/n} \left[\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right]}, \quad k = 0, 1, 2, \ldots, n-1zk=r1/n[cos(nθ+2kπ)+isin(nθ+2kπ)],k=0,1,2,…,n−1
🔹 Key Ideas
✅ There are exactly nnn distinct nnnth roots of any nonzero complex number.
✅ Their moduli are equal (r1/nr^{1/n}r1/n).
✅ Their arguments are equally spaced by 2πn\frac{2\pi}{n}n2π.
✅ Geometrically, they form the vertices of a regular n-sided polygon on the Argand plane, centered at the origin.
🔹 Example 1 — Cube Roots of Unity
Find all cube roots of 1.
We can write: 1=1(cos0+isin0)1 = 1(\cos 0 + i\sin 0)1=1(cos0+isin0)
Using the formula: zk=cos(0+2kπ3)+isin(0+2kπ3),k=0,1,2z_k = \cos\left(\frac{0 + 2k\pi}{3}\right) + i\sin\left(\frac{0 + 2k\pi}{3}\right), \quad k=0,1,2zk=cos(30+2kπ)+isin(30+2kπ),k=0,1,2 z0=1,z1=cos2π3+isin2π3=−12+i32,z2=−12−i32z_0 = 1, \quad z_1 = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}, \quad z_2 = -\frac{1}{2} – i\frac{\sqrt{3}}{2}z0=1,z1=cos32π+isin32π=−21+i23,z2=−21−i23
✅ The cube roots of unity are: 1,ω=−12+i32,ω2=−12−i32\boxed{1, \quad \omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}, \quad \omega^2 = -\frac{1}{2} – i\frac{\sqrt{3}}{2}}1,ω=−21+i23,ω2=−21−i23
and satisfy 1+ω+ω2=01 + \omega + \omega^2 = 01+ω+ω2=0.
🔹 Example 2 — Square Roots of iii
Let z=i=cosπ2+isinπ2z = i = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}z=i=cos2π+isin2π
We want zk=iz_k = \sqrt{i}zk=i, i.e., n=2n = 2n=2. zk=(cosπ/2+2kπ2+isinπ/2+2kπ2),k=0,1z_k = (\cos\frac{\pi/2 + 2k\pi}{2} + i\sin\frac{\pi/2 + 2k\pi}{2}), \quad k = 0, 1zk=(cos2π/2+2kπ+isin2π/2+2kπ),k=0,1 z0=cosπ4+isinπ4=1+i2z_0 = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{1 + i}{\sqrt{2}}z0=cos4π+isin4π=21+i z1=cos5π4+isin5π4=−1−i2z_1 = \cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4} = \frac{-1 – i}{\sqrt{2}}z1=cos45π+isin45π=2−1−i
✅ The two square roots of iii are: 1+i2, −1−i2\boxed{\frac{1 + i}{\sqrt{2}}, \ \frac{-1 – i}{\sqrt{2}}}21+i, 2−1−i
🔹 Example 3 — Cube Roots of 8(cos60°+isin60°)8(\cos 60° + i\sin 60°)8(cos60°+isin60°)
Here r=8r = 8r=8, θ=60°=π3\theta = 60° = \frac{\pi}{3}θ=60°=3π, n=3n = 3n=3. zk=81/3[cos(π/3+2kπ3)+isin(π/3+2kπ3)]z_k = 8^{1/3}\left[\cos\left(\frac{\pi/3 + 2k\pi}{3}\right) + i\sin\left(\frac{\pi/3 + 2k\pi}{3}\right)\right]zk=81/3[cos(3π/3+2kπ)+isin(3π/3+2kπ)] r1/3=2r^{1/3} = 2r1/3=2
So: z0=2(cosπ9+isinπ9)z_0 = 2(\cos\frac{\pi}{9} + i\sin\frac{\pi}{9})z0=2(cos9π+isin9π) z1=2(cos7π9+isin7π9)z_1 = 2(\cos\frac{7\pi}{9} + i\sin\frac{7\pi}{9})z1=2(cos97π+isin97π) z2=2(cos13π9+isin13π9)z_2 = 2(\cos\frac{13\pi}{9} + i\sin\frac{13\pi}{9})z2=2(cos913π+isin913π)
✅ These 3 points form an equilateral triangle centered at the origin.
🌀 4️⃣ Geometric Interpretation
(Illustrate equal-radius points equally spaced around the origin.)
- All roots lie on a circle of radius r1/nr^{1/n}r1/n.
- They are symmetrically spaced by 2πn\frac{2\pi}{n}n2π.
- The first root corresponds to the principal argument; others are obtained by successive rotations.
Complex roots don’t just exist — they form patterns of perfect symmetry.
🧮 5️⃣ Summary Table
| Type | Expression | Roots Count | Geometry |
|---|---|---|---|
| General nth root | r1/n[cos(θ+2kπn)+isin(θ+2kπn)]r^{1/n}[\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})]r1/n[cos(nθ+2kπ)+isin(nθ+2kπ)] | nnn | Regular n-gon |
| Cube roots of unity | 1,ω,ω21, \omega, \omega^21,ω,ω2 | 3 | Equilateral triangle |
| Square roots of i | 1+i2,−1−i2\frac{1+i}{\sqrt{2}}, \frac{-1-i}{\sqrt{2}}21+i,2−1−i | 2 | Opposite points on circle |
| nth roots of real negative number | Lie symmetrically about real axis | nnn | Symmetric pairings |
🔹 Common Mistakes
- ❌ Forgetting the 2kπ2k\pi2kπ term → only one root found.
- ❌ Using degrees and radians inconsistently.
- ❌ Not checking for distinct arguments within 2π2\pi2π.
- ❌ Forgetting geometric meaning (modulus and equal spacing).
🌟 Why It Matters
The concept of complex roots appears in:
- De Moivre’s theorem, Euler’s formula,
- Polynomials and roots of unity,
- Trigonometric identities,
- Physics & signal processing (phasors, waves),
- STEP/MAT problems involving loci and geometry in the Argand plane.
The roots of a complex number are not random — they form perfect geometric harmony.
📘 Learn Beyond the Formula
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In the Mathematics Elevate Mentorship Program, you’ll:
✅ Derive complex results visually,
✅ Master De Moivre’s theorem intuitively,
✅ Solve STEP/MAT-level problems on loci and symmetry.
🚀 Visualize the Argand plane like never before.
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