Lesson 13: Complex Numbers in Polar Form and De Moivre's Theorem

Complex numbers become vastly more powerful in polar form. Addition is easy in rectangular form a + bi, but multiplication, powers and roots are clumsy there — and effortless in polar form, where multiplying means multiplying lengths and adding angles. De Moivre's theorem turns exponentiation into a single multiplication of the angle, and it is the engine behind roots of complex numbers and much of signal processing.

Modulus and argument

For z = a + bi, the modulus is |z| = sqrt(a^2 + b^2) (the distance from the origin in the complex plane) and the argument arg z is the angle θ from the positive real axis, with cos θ = a/|z|, sin θ = b/|z|. The polar form is z = r(cos θ + i sin θ), often abbreviated r·cis θ. Worked examples. z = 1 − i·sqrt(3): |z| = sqrt(1 + 3) = 2, and the point is in quadrant IV with reference angle π/3, so θ = −π/3 (equivalently 5π/3): z = 2 cis(−π/3). z = −3 + 3i: |z| = 3·sqrt(2), quadrant II, θ = 3π/4. As with polar coordinates in Lesson 8, the quadrant decides the argument — arctan(b/a) alone returns the wrong angle in quadrants II and III.

Multiplication and division in polar form

Multiply: moduli multiply, arguments add. Divide: moduli divide, arguments subtract. Worked example: (2 cis(π/6))·(3 cis(π/4)) = 6 cis(π/6 + π/4) = 6 cis(5π/12). In rectangular form that is 6 cos 75° + 6i sin 75° ≈ 1.55 + 5.80i — try obtaining that by FOIL from the rectangular forms and the polar advantage becomes obvious. Division example: cis(80°)/cis(20°) = cis(60°) = 1/2 + i·(sqrt(3)/2). Geometrically, multiplication by cis θ rotates the plane by θ, and multiplication by a positive real r scales it — every complex multiplication is a rotation-and-scaling, which is why complex numbers run AC circuit analysis and computer graphics.

De Moivre's theorem: powers

De Moivre: (r cis θ)^n = r^n cis(nθ) for any integer n. Worked example: (1 + i)^8. Here r = sqrt(2), θ = π/4, so (1 + i)^8 = (sqrt(2))^8 cis(8 · π/4) = 16 cis(2π) = 16 — a purely real answer from eight complex factors. (Repeated squaring agrees: (1 + i)^2 = 2i, (2i)^2 = −4, (−4)^2 = 16 ✓.) Second example: (1 − i·sqrt(3))^3: r = 2, θ = −π/3, so the cube is 8 cis(−π) = −8. The pattern to internalise: powers are trivial in polar form because the angle — the only hard part in rectangular form — simply scales.

Roots of complex numbers

Every nonzero complex number has exactly n distinct nth roots, evenly spaced around a circle: if z = r cis θ, the roots are r^(1/n) cis((θ + 2πk)/n) for k = 0, 1, ..., n − 1. Worked example — cube roots of −8. Write −8 = 8 cis(π): modulus of each root is 8^(1/3) = 2, arguments (π + 2πk)/3 = π/3, π, 5π/3. The roots: 2 cis(π/3) = 1 + i·sqrt(3); 2 cis(π) = −2; 2 cis(5π/3) = 1 − i·sqrt(3). Verify the first: (1 + i·sqrt(3))^2 = −2 + 2i·sqrt(3), and multiplying by (1 + i·sqrt(3)) again gives −2 − 6 = −8 ✓ (the cross terms cancel). Note the real root −2 is just one of three equals — in the complex world, "the cube root of −8" is a set of three numbers, spaced 120° apart on the circle of radius 2.

Common mistakes

• Argument from arctan without the quadrant. For z = −1 − i, arctan(1) = π/4 but the point is in quadrant III: arg z = −3π/4 (or 5π/4). Draw the point; the picture settles the angle.
• Adding moduli when multiplying. (2 cis θ)(3 cis φ) has modulus 6, not 5: lengths multiply. The same slip in division (subtracting moduli) breaks De Moivre too.
• Stopping at one root. x^3 = −8 has three complex solutions. Reporting only −2 misses the geometry (equal spacing of 2π/n) and two-thirds of the marks.
• Raising to a power in rectangular form. Expanding (1 + i)^8 by binomial theorem is possible but slow and error-prone; three lines of De Moivre do it exactly. Convert first.
• Inconsistent angle units. cis expects one unit throughout the problem: π/6 + π/4 = 5π/12 works in radians; mixing 30° with π/4 corrupts the sum silently.

Exam Experience

1. For z = −3 + 3i, the modulus and (principal) argument are

• (A) |z| = 6, arg z = 3π/4
• (B) |z| = 3·sqrt(2), arg z = 3π/4
• (C) |z| = 3·sqrt(2), arg z = −π/4
• (D) |z| = 18, arg z = 2π/3

Answer: (B). |z| = sqrt(9 + 9) = sqrt(18) = 3·sqrt(2); the point is in quadrant II at 45° from the negative real axis, so arg z = 3π/4. (C) uses the reference angle with the wrong quadrant — the signature error.

2. (3 cis(π/3))·(2 cis(π/6)) equals

• (A) 5 cis(π/2)
• (B) 6 cis(π/18)
• (C) 6 cis(π/2) = 6i
• (D) 6 cis(2π/9)

Answer: (C). Moduli multiply (3 × 2 = 6), arguments add (π/3 + π/6 = π/2), and 6 cis(π/2) = 6i. (A) adds the moduli; (B) and (D) mishandle the angle sum.

3. (1 + i)^4 =

• (A) 4i
• (B) −4
• (C) 4
• (D) −4i

Answer: (B). (1 + i) = sqrt(2) cis(π/4); the fourth power is (sqrt(2))^4 cis(π) = 4·(−1) = −4. Direct check: (1 + i)^2 = 2i and (2i)^2 = −4 ✓.

Constructed response. Let z = 1 − i·sqrt(3). (a) Write z in polar form r cis θ with −π < θ ≤ π. (b) Use De Moivre's theorem to find z^6, showing that it is a real number. (c) Find all three cube roots of −8 in rectangular form, and verify one of the non-real roots by cubing it.

Scoring guide. (a) 2 points: r = sqrt(1 + 3) = 2 — 1 point; quadrant IV with reference angle π/3 gives θ = −π/3 — 1 point. (b) 2 points: z^6 = 2^6 cis(6 × (−π/3)) = 64 cis(−2π) — 1 point; cis(−2π) = 1, so z^6 = 64, real — 1 point. (c) 3 points: roots have modulus 2 and arguments π/3, π, 5π/3 — 1 point; rectangular forms 1 + i·sqrt(3), −2, 1 − i·sqrt(3) — 1 point; verification: (1 + i·sqrt(3))^2 = −2 + 2i·sqrt(3), times (1 + i·sqrt(3)) gives −2 − 2i·sqrt(3) + 2i·sqrt(3) − 6 = −8 — 1 point.

Going deeper: roots of unity, Euler's formula and modulus laws

The nth roots of 1 — the roots of unity — are the vertices of a regular n-gon on the unit circle: for n = 4 they are 1, i, −1, −i (arguments 0, π/2, π, 3π/2 by De Moivre). Their sum is 0 (the polygon's centroid is the origin — a fact provable in one line by the geometric series formula of Lesson 10, since the roots form a geometric progression with ratio cis(2π/n)). Roots of unity run the discrete Fourier transform, hence audio processing, image compression and much of modern signal mathematics.

Modulus laws mirror the polar rules: |zw| = |z||w| and |z/w| = |z|/|w|, provable directly from polar form. Rectangular check with z = 1 − i·sqrt(3) (|z| = 2) and w = 1 + i (|w| = sqrt(2)): zw = (1 + sqrt(3)) + (1 − sqrt(3))i, and |zw|^2 = (1 + sqrt(3))^2 + (1 − sqrt(3))^2 = (4 + 2·sqrt(3)) + (4 − 2·sqrt(3)) = 8, so |zw| = 2·sqrt(2) = |z||w| ✓.

Euler's formula compresses everything: e^(iθ) = cos θ + i sin θ, so polar form is simply z = r·e^(iθ), multiplication becomes exponent addition (the polar rules *are* the exponent laws), and De Moivre becomes (e^(iθ))^n = e^(inθ). Its most famous special case, e^(iπ) + 1 = 0, links e, π, i, 1 and 0 in one line — routinely called the most beautiful equation in mathematics, and, after this lesson, one you can actually read.

Workshop: multiplication two ways, and square roots of complex numbers

Rectangular versus polar, head to head. Compute (3 + 4i)(1 − 2i). Rectangular FOIL: 3 − 6i + 4i − 8i^2 = 11 − 2i. Polar check: |3 + 4i| = 5 with argument arctan(4/3) ≈ 53.13°; |1 − 2i| = sqrt(5) with argument ≈ −63.43°. The product must have modulus 5·sqrt(5) = sqrt(125) and argument 53.13° − 63.43° = −10.3°. Indeed |11 − 2i| = sqrt(121 + 4) = sqrt(125) ✓ and arctan(−2/11) ≈ −10.3° ✓. The two forms are two audits of one computation — in an exam, run both when the stakes justify it.

Square roots of a complex number. Solve w^2 = 3 + 4i with w = a + bi: a^2 − b^2 = 3 and 2ab = 4, so b = 2/a and a^2 − 4/a^2 = 3 → a^4 − 3a^2 − 4 = 0 → (a^2 − 4)(a^2 + 1) = 0 → a = ±2 (real a), b = ±1 with matching signs: w = 2 + i or w = −2 − i. Verify: (2 + i)^2 = 4 + 4i + i^2 = 3 + 4i ✓. De Moivre finds these roots faster (halve the argument of 5 cis(53.13°): sqrt(5) cis(26.57°), whose rectangular form is exactly 2 + i) — the rectangular method is shown here to prove the two approaches agree, and because "solve a^2 − b^2, 2ab" is the fallback when angles are not special.

Final drills: conjugates and a power

Conjugates multiply to the modulus squared. For z = 3 − 4i, the conjugate is z̄ = 3 + 4i and z·z̄ = 9 + 16 = 25 = |z|^2 — the cross terms cancel identically (a + bi)(a − bi) = a^2 + b^2. This is the trick that divides complex numbers (multiply top and bottom by the conjugate to real-ise the denominator) and it is the same rationalising move as Lesson 9's surds, one level up. A De Moivre sprint: (2 cis(π/3))^3 = 8 cis(π) = −8: three equal factors, each a 60° rotation, land on the negative real axis. Rotations compose by addition — that single sentence is the whole lesson, and complex arithmetic is bookkeeping built on top of it.