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.
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), …
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.
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.