Complex Numbers: The Map

A2 · P3 · 2 min

Complex numbers extend the real numbers with a single new symbol, ii, whose square is −1-1. With it every polynomial equation has solutions, and a number becomes a point in a plane rather than a point on a line. This unit is a sequence of skills that build on each other; use this page to see how they fit.

The notes in order

  1. Introduction: i2=−1i^2 = -1, real and imaginary parts, adding, subtracting, multiplying, the conjugate, and equating parts.
  2. Division: multiply by the conjugate; solving equations with complex coefficients.
  3. Square roots: a+bi\sqrt{a + bi} by equating parts.
  4. Complex roots of polynomials: conjugate pairs and factorising cubics and quartics.
  5. The complex plane: Argand diagram, modulus ∣z∣|z| and argument arg⁡z\arg z.
  6. Geometry of operations: what conjugation, addition and multiplication do to a point.
  7. Polar form: r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta) and multiplying or dividing by adding or subtracting arguments.
  8. Exponential form: reiθre^{i\theta}.
  9. Loci: circles, perpendicular bisectors and half-lines from ∣z−a∣=k|z - a| = k, ∣z−a∣=∣z−b∣|z - a| = |z - b|, arg⁡(z−a)=α\arg(z - a) = \alpha.

The notation at a glance

Key result

For z=x+iyz = x + iy with x,yx, y real:

SymbolMeaningValue
Re⁡z\operatorname{Re} zreal partxx
Im⁡z\operatorname{Im} zimaginary partyy (not iyiy)
z∗z^*conjugatex−iyx - iy
∣z∣\lvert z \rvertmodulusx2+y2\sqrt{x^2 + y^2}
arg⁡z\arg zargumentangle from the positive real axis, usually −π<arg⁡z≤π-\pi < \arg z \leq \pi

Two complex numbers are equal if and only if both their real parts and their imaginary parts are equal.

The three forms

z=x+iy=r(cos⁡θ+isin⁡θ)=reiθ,r=∣z∣, θ=arg⁡z.z = x + iy = r(\cos\theta + i\sin\theta) = re^{i\theta}, \qquad r = |z|,\ \theta = \arg z.

Cartesian form is for adding and subtracting; polar and exponential forms are for multiplying, dividing and reading off geometry.

A tour through the forms

Let z=1+i3z = 1 + i\sqrt{3}. Find z∗z^*, ∣z∣|z|, arg⁡z\arg z, and write zz in polar and exponential form. Then find z2z^2 two ways.

Solution

z∗=1−i3z^* = 1 - i\sqrt{3}. ∣z∣=1+3=2|z| = \sqrt{1 + 3} = 2. arg⁡z=tan⁡−13=π3\arg z = \tan^{-1}\sqrt{3} = \tfrac{\pi}{3} (first quadrant).

z=2(cos⁡π3+isin⁡π3)=2eiπ/3z = 2\left(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}\right) = 2e^{i\pi/3}.

Cartesian: z2=1+2i3−3=−2+2i3z^2 = 1 + 2i\sqrt{3} - 3 = -2 + 2i\sqrt{3}. Polar: z2=4e2iπ/3=4(−12+i32)=−2+2i3z^2 = 4e^{2i\pi/3} = 4\left(-\tfrac{1}{2} + i\tfrac{\sqrt{3}}{2}\right) = -2 + 2i\sqrt{3} ✓.

Exam tip

A typical P3 question walks through several of these skills in order: solve a quadratic with complex roots, plot them, find modulus and argument, then sketch a locus. Each part is short; the marks come from doing the standard step cleanly and showing the working the syllabus insists on (multiplication, division and square roots in full).

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