# Chapter 28: Complex Numbers

> From i to Euler's formula: arithmetic, conjugates and moduli, the complex plane, De Moivre powers, all n roots of zⁿ = w, logarithms of complex numbers — and why e^(iπ) + 1 = 0 is the jewel of the subject.

**Canonical:** https://calculus.kitzo.work/chapter/complex-numbers  
**Level:** Advanced  
**Estimated time:** 45 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Complex arithmetic
- |z| and conjugates
- Euler's formula
- Polar form
- De Moivre's theorem
- Solving zⁿ = w
- Euler's identity

**Why this matters**

Complex numbers fuse algebra with geometry — multiplication is rotation, exponentiation is rotation magnified, and Euler's formula e^(iθ) = cos θ + i sin θ turns trigonometric identities into one-line computations. They are the language of AC circuits, quantum mechanics, signal processing — and secretly the fastest way through a lot of calculus.

## 1 · The basics — i, arithmetic, conjugates, |z|

Define i by **i² = −1**. Write z = a + bi with real part a and imaginary part b. Add and subtract componentwise; multiply by FOIL using i² = −1; divide by multiplying numerator and denominator by the conjugate. The **conjugate** is z̄ = a − bi; the **modulus** is |z| = √(a² + b²); and **z·z̄ = |z|²**, always a nonnegative real number.

**Worked example**

**Example 1 — Compute (2 + 3i)/(1 − i).**

Step 1: multiply by the conjugate (1 + i)/(1 + i).

Step 2: numerator (2 + 3i)(1 + i) = 2 + 2i + 3i + 3i² = −1 + 5i.

Step 3: denominator (1 − i)(1 + i) = 1 − i² = 2. Result: −½ + (5/2)i.

## 2 · Euler's formula — e^(iθ) = cos θ + i sin θ

Substitute iθ into the series eˣ = Σ xⁿ/n!: the even-powered terms rebuild the cosine series, the odd-powered terms rebuild sine, and the i factors tag them apart. Hence **e^(iθ) = cos θ + i sin θ**. Every rule of ordinary exponentials carries over, e.g. e^(iθ)·e^(iφ) = e^(i(θ+φ)) — because they are the same series.

## 3 · The complex plane and polar form z = re^(iθ)

Plot z = a + bi as (a, b) on the **complex plane** (horizontal real axis, vertical imaginary). The **modulus** r = |z| = √(a² + b²) is the distance from the origin; the **argument** θ = arg z (principal value, −π < θ ≤ π) is the angle from the positive x-axis. Convert with a = r cos θ, b = r sin θ, and write **z = r e^(iθ)**. Polar form turns multiplication into "multiply the radii, add the angles".

## 4 · De Moivre's theorem — big powers with ease

Raising to the n-th power: **(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ**, i.e. zⁿ = rⁿ e^(inθ). Tedious powers become instant. Compute (1 + i)¹⁰: |1 + i| = √2 and arg = π/4, so (1 + i)¹⁰ = (√2)¹⁰ e^(i10π/4) = 32·e^(i5π/2) = 32i. No binomial expansion needed.

## 5 · Solving zⁿ = w — all n roots

Write the target in polar form w = ρ e^(iφ). Then **zₖ = ρ^(1/n) · e^(i(φ + 2kπ)/n)** for k = 0, 1, …, n − 1 — exactly n roots, evenly spaced on a circle of radius ρ^(1/n), starting at angle φ/n. For w = 1 these are the **n-th roots of unity**: 1, e^(2πi/n), e^(4πi/n), …

**Example 2 — Solve z³ = 8i.**

Step 1: 8i = 8 e^(iπ/2), so ρ = 8, φ = π/2.

Step 2: radius 8^(1/3) = 2; angles (π/2 + 2kπ)/3 = π/6, 5π/6, 3π/2.

Step 3: roots: 2e^(iπ/6) = √3 + i; 2e^(i5π/6) = −√3 + i; 2e^(i3π/2) = −2i. Each cubes to 8i — a full check.

## 6 · e^z = w, trigonometric series, Euler's identity

Solve **e^z = w** by writing w = ρ e^(iφ): then z = ln ρ + i(φ + 2kπ), k any integer — infinitely many logarithms, spaced by 2πi, because e^z is periodic with period 2πi. Trigonometric series appear as real parts: cos nθ = Re(e^(inθ)) = Re((cos θ + i sin θ)ⁿ), which expands into polynomials in cos θ (e.g. cos 3θ = 4cos³θ − 3cos θ).

**Euler's identity**: **e^(iπ) + 1 = 0** — e, i, π, 1 and 0 in one equation. And the circle closes back to series: e^z converges for all complex z, and z = iθ reproduces the cosine and sine series; all of Taylor theory extends verbatim to complex input.

**Watch out**

arg z is multi-valued: adding 2π gives the same z — quote the principal value unless asked otherwise, and never forget the + 2kπ term when listing roots of zⁿ = w or logarithms of w. Also e^z is not one-to-one, and i is not "the square root of negative one" in any formula that quietly assumes real inputs.

**Chapter formula sheet**

|z| = √(a² + b²) · z·z̄ = |z|² · conj(z₁z₂) = z̄₁z̄₂

e^(iθ) = cos θ + i sin θ · (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ

z = re^(iθ), r = |z|, θ = arg z (principal −π < θ ≤ π)

zⁿ = w ⇒ zₖ = ρ^(1/n) e^(i(φ + 2kπ)/n), k = 0, …, n − 1

e^z = w ⇒ z = ln ρ + i(φ + 2kπ) · e^(iπ) + 1 = 0

## Common mistakes

  - Applying √(ab) = √a·√b to negative numbers — it fails there.

  - Reporting one root of zⁿ = w instead of all n (the + 2kπ omission).

  - Reading θ = arctan(b/a) for (−3, −4) as a first-quadrant angle — always fix the quadrant.

  - Claiming |z₁ + z₂| = |z₁| + |z₂| — the triangle inequality is ≤.

  - Mixing degrees and radians inside e^(iθ) — it is always radians.

## Practice exercises

1. What is |3 + 4i| ?

- A. 25
- B. 5
- C. 7
- D. 1

2. Compute |6 + 8i| .

3. Euler's identity states e^(iπ) = … (include the sign).

4. Compute (1 + i)^8 .

5. z³ = 1 has how many solutions in the complex numbers?

- A. 1
- B. 2
- C. 3
- D. 6

6. Find arg(1 + i) in radians (principal value, ≈ 3 decimals).

7. Write the conjugate of z = 3 + 4i without spaces, e.g. like 3-4i .

8. Euler's formula says e^(iθ) equals:

- A. cos θ + i sin θ
- B. cos θ − i sin θ
- C. sin θ + i cos θ
- D. i sin θ

---

Practice interactively: https://calculus.kitzo.work/chapter/complex-numbers

Citation: CalculusKitzo (https://calculus.kitzo.work/). Independent study notes inspired by the standard calculus syllabus; not affiliated with Princeton University Press. See https://calculus.kitzo.work/disclaimer