Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)**
Published: October 2025
Category: Complex Numbers | Trigonometry & Geometry
🔹 Introduction
Complex numbers are not just algebraic curiosities — they form a geometric system where algebra, trigonometry, and exponential functions meet beautifully.
One result stands above the rest in uniting them: De Moivre’s Theorem.
It transforms multiplication and exponentiation in complex numbers into simple operations on angles — turning geometry into algebra.
🧭 1️⃣ Polar Form of a Complex Number
Any complex number can be written as: z=r(cosθ+isinθ)z = r(\cos\theta + i\sin\theta)z=r(cosθ+isinθ)
where
- r=∣z∣=x2+y2r = |z| = \sqrt{x^2 + y^2}r=∣z∣=x2+y2 is the modulus,
- θ=arg(z)=tan−1 (yx)\theta = \arg(z) = \tan^{-1}\!\left(\frac{y}{x}\right)θ=arg(z)=tan−1(xy) is the argument.
In Euler’s form, z=reiθ.z = re^{i\theta}.z=reiθ.
⚡️ 2️⃣ De Moivre’s Theorem
If z=r(cosθ+isinθ)z = r(\cos\theta + i\sin\theta)z=r(cosθ+isinθ), then for any integer nnn: zn=rn(cosnθ+isinnθ)\boxed{z^n = r^n(\cos n\theta + i\sin n\theta)}zn=rn(cosnθ+isinnθ)
🔹 Proof (using induction)
Base case n = 1: true trivially.
Assume true for n = k: zk=rk(coskθ+isinkθ)z^k = r^k(\cos k\theta + i\sin k\theta)zk=rk(coskθ+isinkθ)
Then zk+1=zk⋅z=rk+1 [(coskθ+isinkθ)(cosθ+isinθ)]z^{k+1} = z^k · z = r^{k+1}\!\left[(\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)\right]zk+1=zk⋅z=rk+1[(coskθ+isinkθ)(cosθ+isinθ)]
Using trig identities: cos(kθ+θ)+isin(kθ+θ)\cos(k\theta + \theta) + i\sin(k\theta + \theta)cos(kθ+θ)+isin(kθ+θ) =cos((k+1)θ)+isin((k+1)θ)= \cos((k+1)\theta) + i\sin((k+1)\theta)=cos((k+1)θ)+isin((k+1)θ)
✅ Hence true for n = k + 1 → proved by induction.
🔹 Using Euler’s Form
z=reiθ⇒zn=(reiθ)n=rneinθ=rn(cosnθ+isinnθ)z = re^{i\theta} \Rightarrow z^n = (re^{i\theta})^n = r^n e^{in\theta} = r^n(\cos n\theta + i\sin n\theta)z=reiθ⇒zn=(reiθ)n=rneinθ=rn(cosnθ+isinnθ)
✅ Elegant one-line proof using Euler’s identity eiθ=cosθ+isinθ.e^{i\theta} = \cos\theta + i\sin\theta.eiθ=cosθ+isinθ.
🎯 3️⃣ Applications of De Moivre’s Theorem
✅ (1) Powers of Complex Numbers
Simplifies znz^nzn directly without repeated multiplication.
Example: (1+i)5(1 + i)^5(1+i)5 r=2, θ=45°=π4r = \sqrt{2},\ \theta = 45° = \frac{\pi}{4}r=2, θ=45°=4π z5=(2)5[cos(5π/4)+isin(5π/4)]=42(−12−i12)=−4−4iz^5 = (\sqrt{2})^5 [\cos(5\pi/4) + i\sin(5\pi/4)] = 4\sqrt{2}(-\frac{1}{\sqrt{2}} – i\frac{1}{\sqrt{2}}) = -4 – 4iz5=(2)5[cos(5π/4)+isin(5π/4)]=42(−21−i21)=−4−4i
✅ So (1+i)5=−4−4i(1+i)^5 = -4 – 4i(1+i)5=−4−4i
✅ (2) Finding Roots of Complex Numbers
De Moivre’s theorem gives all nthn^{\text{th}}nth roots: z1/n=r1/n [cos (θ+2kπn)+isin (θ+2kπn)], k=0,1,…,n−1z^{1/n} = r^{1/n}\!\left[\cos\!\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\!\left(\frac{\theta + 2k\pi}{n}\right)\right], \ k = 0,1,…,n-1z1/n=r1/n[cos(nθ+2kπ)+isin(nθ+2kπ)], k=0,1,…,n−1
Example: Cube roots of 1: 1, ω=−12+i32, ω2=−12−i321,\ \omega = -\tfrac{1}{2} + i\tfrac{\sqrt{3}}{2},\ \omega^2 = -\tfrac{1}{2} – i\tfrac{\sqrt{3}}{2}1, ω=−21+i23, ω2=−21−i23
✅ Form equilateral triangle on Argand plane.
✅ (3) Trigonometric Identities
Using (cosθ+isinθ)n=cosnθ+isinnθ(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta(cosθ+isinθ)n=cosnθ+isinnθ,
equating real and imaginary parts gives: cosnθ=ℜ[(cosθ+isinθ)n]\cos n\theta = \Re[(\cos\theta + i\sin\theta)^n]cosnθ=ℜ[(cosθ+isinθ)n] sinnθ=ℑ[(cosθ+isinθ)n]\sin n\theta = \Im[(\cos\theta + i\sin\theta)^n]sinnθ=ℑ[(cosθ+isinθ)n]
✅ Generates multiple-angle and reduction formulas.
✅ (4) Geometric Interpretation
Each power of zzz multiplies the modulus and rotates the angle.
- r→rnr → r^nr→rn: scaling
- θ→nθ\theta → n\thetaθ→nθ: rotation
Raising a complex number to a power is a rotation + scaling transformation in the Argand plane.
🧩 4️⃣ Example Set
| Example | Input | Result |
|---|---|---|
| (1−i)4(1 – i)^4(1−i)4 | r=√2, θ=−45°r=√2,\ θ=−45°r=√2, θ=−45° | (√2)4(cos−180°+isin−180°)=4(−1)=−4(√2)^4(\cos −180° + i\sin −180°) = 4(−1) = −4(√2)4(cos−180°+isin−180°)=4(−1)=−4 |
| (i)6(i)^6(i)6 | r=1, θ=90°r=1,\ θ=90°r=1, θ=90° | 1×(cos540°+isin540°)=cos180°=−11 × (\cos 540° + i\sin 540°) = \cos 180° = −11×(cos540°+isin540°)=cos180°=−1 |
| (cos20°+isin20°)9(\cos 20° + i\sin 20°)^9(cos20°+isin20°)9 | — | cos180°+isin180°=−1\cos 180° + i\sin 180° = −1cos180°+isin180°=−1 |
🧮 5️⃣ Summary Table
| Concept | Formula | Interpretation |
|---|---|---|
| Polar form | z=r(cosθ+isinθ)z = r(\cos θ + i \sin θ)z=r(cosθ+isinθ) | Position on Argand plane |
| De Moivre’s Theorem | zn=rn(cosnθ+isinnθ)z^n = r^n(\cos nθ + i \sin nθ)zn=rn(cosnθ+isinnθ) | Multiply modulus, rotate angle |
| Roots formula | zk=r1/n[cos((θ+2kπ)/n)+isin((θ+2kπ)/n)]z_k = r^{1/n}[\cos((θ + 2kπ)/n) + i\sin((θ + 2kπ)/n)]zk=r1/n[cos((θ+2kπ)/n)+isin((θ+2kπ)/n)] | n equally spaced roots |
| Geometric effect | Scaling + rotation | Transformation in plane |
🔹 Common Mistakes
- ❌ Forgetting that θ is in radians when using π.
- ❌ Ignoring the 2kπ2kπ2kπ term → missing other roots.
- ❌ Not taking the principal argument in (–π, π].
- ❌ Sign errors in sin/cos for quadrant angles.
🌟 Why It Matters
De Moivre’s Theorem is the foundation for:
- Roots of unity and factorization of polynomials
- Trigonometric identity derivations
- Complex exponentials in physics and signals
- STEP/MAT geometry and rotation questions
It shows how raising a number to a power creates perfect symmetry in the Argand plane.
📘 Learn Beyond Formula
At Math By Rishabh, you learn why De Moivre’s Theorem works — not just how to apply it.
In the Mathematics Elevate Mentorship Program, you’ll:
✅ Visualize complex powers as rotations,
✅ Derive identities from first principles,
✅ Master IB HL / A Level / STEP complex geometry.
🚀 See complex numbers as geometry in motion.
👉 Book your personalized mentorship session now at MathByRishabh.com
