Complex Number Algebra

01 / UNDERSTAND & APPLY

1. Two real coordinates form one complex number

A complex number z = x + iy has real coordinates x and y, with i² = −1. Its real part is Re z = x and its imaginary part is Im z = y, a real number. The term iy is the imaginary component, but Im z itself is y rather than iy. Two complex numbers are equal exactly when both corresponding real coordinates agree.

Addition and subtraction act coordinate by coordinate. Multiplication uses the distributive law and i² = −1: (x + iy)(u + iv) = (xu − yv) + i(xv + yu). Powers of i repeat with period four, so a high integer power can be reduced by its exponent modulo four. Complex arithmetic extends real arithmetic, but it does not impose a natural order compatible with the field operations.

The conjugate z̄ = x − iy reflects z across the real axis. It obeys z + z̄ = 2x, z − z̄ = 2iy, and zz̄ = x² + y². Conjugation respects sums, products and quotients with nonzero denominator. These identities are useful both for arithmetic and for converting a condition involving z and z̄ into two real equations.

The modulus |z| = √(x² + y²) is a nonnegative real number. It vanishes only at z = 0. Do not confuse z² with |z|²: the latter equals zz̄ and is real, while z² need not be. For instance, i² = −1 but |i|² = 1. An inequality involving |z| is meaningful as a real inequality; an unqualified statement z > 0 is not the standard order relation of complex arithmetic.

02 / UNDERSTAND & APPLY

2. Division and equations involving a conjugate

For w ≠ 0, write z/w = z w̄/|w|². The real positive denominator makes rectangular simplification convenient. Division by a nonzero imaginary number is perfectly valid; rationalizing the denominator is a representation choice, not a mathematical requirement that denominators must be real. The actual restriction is only w ≠ 0.

An equation involving z̄ is often easiest with z = x + iy. Conjugation is not multiplication by a fixed complex constant, so do not treat z and z̄ as interchangeable unknowns or apply a polynomial formula blindly. Separate real and imaginary parts and solve the resulting real system.

The modulus is multiplicative: |zw| = |z||w| and |z/w| = |z|/|w| for w ≠ 0. This follows by squaring and using conjugates. The triangle inequality |z + w| ≤ |z| + |w| reflects the fact that a direct path is no longer than a broken path. Equality holds when nonzero z and w point in the same direction; zero-vector cases also give equality.

03 / UNDERSTAND & APPLY

3. Modulus and angle are a second coordinate system

For nonzero z, write z = r(cos θ + i sin θ), where r = |z| > 0 and θ is an argument. All arguments differ by integer multiples of 2π. We use the principal argument Arg z in (−π,π]; another convention is possible, but it must be stated and used consistently. Zero has modulus zero and no defined argument.

Computing θ from tan θ = y/x is not enough because tangent cannot distinguish opposite quadrants. Use the signs of x and y, or equivalent sine and cosine conditions, to choose the correct direction. If x = 0, handle the positive or negative imaginary axis directly rather than dividing by zero.

Multiplication multiplies moduli and adds arguments modulo 2π. Division divides moduli and subtracts arguments modulo 2π, with nonzero divisor. This follows from the trigonometric addition formulas. Principal arguments do not always add as ordinary numbers: after adding, the result may need to be shifted by 2π into the principal interval.

De Moivre’s formula gives [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) for integer n, with z ≠ 0 when n is negative. It is especially useful for large powers. Applying a fractional exponent as though it produced only one answer can lose roots; the next sections keep all possible arguments explicitly.

04 / UNDERSTAND & APPLY

4. Distances, circles and the vector hidden in a difference

On the Argand plane, z = x + iy corresponds to the point (x,y). The difference z − a is the directed displacement from a to z, and |z − a| is their Euclidean distance. Therefore |z − a| = R is a circle of center a and radius R when R > 0, a single point when R = 0, and empty when R < 0.

The locus |z − a| = |z − b| for distinct a and b is the perpendicular bisector of the segment joining them. Squaring cancels the x² + y² terms and leaves a linear equation. For a fixed ratio k > 0 with k ≠ 1, |z − a| = k|z − b| is an Apollonius circle when a ≠ b. The k = 1 case is the line, so it must not be forced into a circle formula with a zero denominator.

Expanding a modulus equation is safe after checking that both sides are nonnegative. Squaring an equation with a negative prescribed right side can introduce an entire false locus. Likewise, if the original condition is a quotient |(z − a)/(z − b)| = k, the point z = b is excluded before multiplying through.

A useful diagram marks the fixed points a and b and a moving point z, then draws the two segments whose lengths are compared. This makes the perpendicular-bisector case obvious and gives an intuitive reason a distance ratio need not describe a straight line.

05 / UNDERSTAND & APPLY

5. Ratios encode alignment and oriented angles

For distinct a and b and a moving point z ≠ a, the ratio (z − b)/(z − a) compares two nonzero displacement directions when z ≠ b as well. A real ratio means the directions are parallel or antiparallel, hence the three points are collinear. A purely imaginary nonzero ratio means the directions are perpendicular. Zero must be excluded in the perpendicular claim because a zero displacement has no direction.

The argument of a quotient is the difference of arguments modulo 2π. An equation prescribing a principal argument selects an oriented angle and often only part of a familiar locus. Squaring or retaining only the real-part condition can accidentally include the opposite arc. Preserve the sign of the imaginary part and any excluded endpoints.

For example, an angle of π/2 subtended by a fixed segment suggests the circle with that segment as diameter. But the positive versus negative oriented angle selects a particular semicircle. Algebra can establish both the circle equation and the correct half-plane, making the diagram precise.

For collinearity, an alternative is Im[(z − a)conj(b − a)] = 0, which avoids dividing by z − a. For perpendicular nonzero displacements u,v, Re(u v̄) = 0 is the dot-product condition. These formulas connect complex geometry to familiar Cartesian vector tests.

06 / UNDERSTAND & APPLY

6. Multiplication is a geometric transformation

The map z ↦ αz + β, with α ≠ 0, rotates by an argument of α, scales distances by |α|, and then translates by β. To rotate about a center c rather than the origin, first subtract c, rotate the displacement, and then add c back: z ↦ c + e^(iθ)(z − c). Here e^(iθ) is shorthand for cos θ + i sin θ.

Multiplication by i gives a quarter-turn counterclockwise, because i(x + iy) = −y + ix. Multiplication by −1 gives a half-turn. Conjugation is different: it reflects across the real axis and reverses orientation, so it cannot be represented by multiplication by one fixed complex number for every z.

The triangle inequality gives useful bounds on distances. For fixed a and b, |z − a| + |z − b| ≥ |a − b|, with equality for z on the segment joining them. The reverse triangle inequality gives ||z − a| − |z − b|| ≤ |a − b|. These bounds can show that a proposed locus is empty before any lengthy algebra.

For a circle |z − c| = R, the distance to a fixed point a ranges between ||a − c| − R| and |a − c| + R. The extremes lie on the line through a and c when a ≠ c. This is a geometric optimization result that avoids differentiation and keeps equality conditions visible.

07 / UNDERSTAND & APPLY

7. Find every root before selecting a principal one

For w ≠ 0 with polar form R(cos φ + i sin φ), solve zⁿ = w with n a positive integer. The modulus of z must be R^(1/n). Its argument satisfies nθ = φ + 2πk, so θ = (φ + 2πk)/n. Taking k = 0,…,n − 1 gives exactly n distinct roots, equally spaced around a circle centered at the origin.

Changing the chosen argument φ by 2π merely permutes the full root set. Taking only θ = φ/n gives one root, not all roots. A principal nth-root convention selects one value, but an equation zⁿ = w normally asks for the whole set unless explicitly restricted. When w = 0, the only distinct solution is z = 0, with multiplicity n in the polynomial.

For square roots of a rectangular number, a direct method is useful: set z = u + iv, equate u² − v² to the real part and 2uv to the imaginary part, and use u² + v² = |w|. The sign relation 2uv fixes which pair of square-root signs belongs together. Independent ± choices would introduce false roots.

08 / UNDERSTAND & APPLY

8. Unity roots form a regular polygon and a cyclic pattern

The nth roots of unity solve zⁿ = 1. With ζ = cos(2π/n) + i sin(2π/n), they are 1,ζ,ζ²,…,ζ^(n−1). Their moduli are one and their arguments are equally spaced by 2π/n, so for n ≥ 3 they form a regular n-gon. The cases n = 1 and n = 2 are a point and two opposite points rather than ordinary polygons.

Multiplying two nth roots adds their exponents modulo n, and taking an inverse negates the exponent. Since they lie on the unit circle, their conjugates equal their reciprocals. A primitive nth root has exact multiplicative order n: its powers visit all n roots before returning to 1.

The order of ζᵏ is n/gcd(n,k). Indeed, (ζᵏ)ʳ = 1 exactly when n divides kr; the smallest positive such r is n/gcd(n,k). Therefore ζᵏ is primitive precisely when gcd(n,k) = 1. The root 1 is not primitive for n > 1, and not every nth root is primitive.

For cube roots, let ω = −1/2 + i√3/2. Then ω³ = 1 but ω ≠ 1, and 1 + ω + ω² = 0. Also ω̄ = ω² and ω + ω² = −1. Reduce high powers modulo three before simplifying an expression. The equation ω³ = 1 alone would also permit ω = 1, so the nontrivial-root condition matters.

09 / UNDERSTAND & APPLY

9. Geometric sums select multiples of n

For n > 1, the sum of all nth roots is zero. Apply the finite geometric-sum identity to 1 + ζ + ⋯ + ζ^(n−1): multiplying by 1 − ζ gives 1 − ζⁿ = 0, and 1 − ζ is nonzero. The polygon’s symmetry gives a second explanation: its position vectors cancel around the center.

More generally, for any integer m, Σₖ₌₀^(n−1)ζ^(mk) equals n if n divides m and zero otherwise. If n divides m, every term is 1. Otherwise ζᵐ ≠ 1 and the geometric-sum numerator is 1 − ζ^(mn) = 0. This roots-of-unity filter is useful for selecting exponents in a polynomial that belong to a fixed residue class.

To select coefficients whose exponents are divisible by n, evaluate a polynomial P at all nth roots and average: (1/n)ΣP(ζᵏ). Expanding P and applying the power-sum identity removes all other exponents. This works for finite polynomials without convergence issues. For a different residue class, an extra weighting factor is needed; do not assume the unweighted average selects every desired class.

Pairs of reciprocal roots can simplify rational sums. For z ≠ 0,1, 1/(1 − z) + 1/(1 − 1/z) = 1. On the unit circle, pairing a root with its conjugate uses this identity. Check separately any self-paired root such as −1 and never include the excluded root 1 in a denominator 1 − z.

10 / UNDERSTAND & APPLY

10. Original practice: algebra and plane geometry

Choose rectangular coordinates for real/imaginary equations, polar form for powers and rotations, and distance language for modulus loci. A good representation can replace pages of expansion.

11 / UNDERSTAND & APPLY

11. Original practice: all roots and unity sums

For a root equation, count distinct answers and verify the intended root set. For a unity sum, decide whether the exponent is divisible by n before applying a geometric formula.

12 / UNDERSTAND & APPLY

12. Keep the representation and its restrictions together

The algebraic form x + iy makes equality a pair of real equations. Polar form makes multiplication and roots a problem of scales and angles. The geometric form |z − a| makes a modulus a distance. Roots of unity turn evenly spaced arguments into finite cyclic algebra. Switching between these forms is the central skill of this family.

The recurring errors are assigning Im z the value iy rather than y, replacing |z|² by z², giving zero an argument, adding principal arguments without reducing modulo 2π, listing only one nth root, and treating every nth root as primitive. In loci, remember excluded denominator points and retain the sign that selects an oriented arc.

Before finalizing a complex answer, check it in another representation. Multiply a quotient back by its denominator, raise roots to the requested power, substitute a simple point into a locus, or verify a rotation preserves distance from its center. For unity sums, compare the algebra with the symmetry of the corresponding regular polygon.

Only claim the geometric conditions actually proved. A real ratio includes both parallel and antiparallel directions; a purely imaginary ratio must be nonzero for perpendicular directions. A circle equation derived from an argument condition may need a half-plane restriction. These small qualifications distinguish a complete solution from a plausible picture.