Complex Numbers

Why the Scary Name Is a Lie

LC Higher Level · Argand diagrams · De Moivre's theorem · conjugate roots · fundamental theorem of algebra

"Complex" is a bad name. It suggests these numbers are harder than the ones you already know. They're not — they're the exact same numbers, plus one new idea: a second axis, running perpendicular to the real number line. Once you see that, most of the syllabus results in this topic stop looking like separate things to memorise and start looking like the same picture, viewed from different angles.

1. One extra axis

A complex number \(z = a + bi\) is just a pair of real numbers, \(a\) and \(b\), written as one object. Plot it on an Argand diagram: the horizontal axis is the familiar real number line, and the vertical axis — measured in units of \(i\) — is new, but it's still just an axis. A complex number is a point in this plane, or equivalently an arrow from the origin out to that point.

Argand diagram with the real and imaginary axes, showing two complex numbers plotted as points: z1 = 3 + 2i and z2 = -2 + i. Re Im z₁ = 3 + 2i z₂ = −2 + i
Two complex numbers plotted as points on the Argand plane.

2. A warm-up on the real line

Before jumping to two dimensions, it's worth noticing that addition and multiplication of ordinary real numbers are already geometric operations — you just might never have thought of them that way.

Addition slides a point along the line: \(2+3=5\) means start at 2 and slide 3 units right. Multiplication scales a point's distance from 0, and multiplying by a negative number also flips it through the origin: \(2\times(-2)=-4\) scales the distance by 2 and flips through 0.

Two number lines: the top shows addition as sliding from 2 to 5; the bottom shows multiplication by minus two as scaling and flipping through zero, from 2 to minus 4. 0 2 5 +3 0 2 −4 ×(−2)
Addition slides; multiplication by a negative number scales and flips through zero — a 180° rotation in disguise.

That "flip through the origin" is secretly a rotation, by exactly 180°. In the complex plane, multiplication does the same two things — scale, and rotate — except the rotation is no longer stuck at 0° or 180°. It can be by any angle at all.

3. Addition is vector addition

Add complex numbers component-by-component: \((a+bi)+(c+di) = (a+c)+(b+d)i\). Geometrically, that's exactly vector addition — draw \(z_1\) and \(z_2\) as arrows from the origin, and their sum is the far corner of the parallelogram they form. Equivalently: walk along \(z_1\), then walk along \(z_2\) from there, and you land in the same place.

Argand diagram showing vector z1 equals four plus i, vector z2 equals one plus three i, and their sum five plus four i as the diagonal of the parallelogram they form. z₁ = 4 + i z₂ = 1 + 3i z₁+z₂ = 5 + 4i
z₁ and z₂ as vectors; their sum is the parallelogram's diagonal.
Try it yourself
Drag z₁ and z₂ to see their sum update in real time.

4. Multiplication is rotate + scale

Write a complex number in modulus–argument (polar) form, \(z = r(\cos\theta + i\sin\theta)\), where \(r=|z|\) is its distance from the origin and \(\theta=\arg z\) is its angle from the positive real axis. Multiplying two complex numbers multiplies their moduli and adds their arguments:

\[ z_1 z_2 = r_1 r_2\big(\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)\big) \]

So multiplying \(z\) by \(w\) rotates \(z\) by \(\arg w\) and scales it by \(|w|\) — the number-line "flip" idea, generalised to any angle: multiplying by \(-1\) is a 180° rotation, multiplying by \(i\) is a 90° rotation, multiplying by any \(w\) is a rotation by \(\arg w\).

Diagram showing a vector z multiplied by w, rotating it by the angle theta (w's argument) and scaling it by w's modulus. θ z wz rotate by θ = arg(w), scale by |w|
Multiplying z by w rotates it by w's argument and scales it by w's modulus.
Try it yourself
Drag the points or sliders to see multiplication rotate and scale z in real time.
Worked algebra — dividing complex numbers

The one algebraic technique this idea leans on is division, which works by multiplying above and below by the conjugate of the denominator so the bottom becomes real:

\[ \frac{3+2i}{1-i} = \frac{(3+2i)(1+i)}{(1-i)(1+i)} = \frac{3+3i+2i+2i^2}{1-i^2} = \frac{1+5i}{2} = \tfrac{1}{2} + \tfrac{5}{2}i \]

The picture explains why division rotates and scales in reverse — but you still need this algebra to get an exact answer.

5. Repeated rotation → De Moivre's theorem

Take a complex number of modulus 1, \(z=\cos\theta+i\sin\theta\), and multiply it by itself repeatedly. The modulus never changes (\(1\times1=1\)), but each multiplication rotates it by \(\theta\) again — so \(z^n\) sits at angle \(n\theta\) around the unit circle. That's De Moivre's theorem:

\[ (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta), \quad n \in \mathbb{Z} \]

Unit circle showing points z to the zero, one, two and three walking round the circle in equal 45 degree steps, illustrating repeated rotation. z⁰=1 z¹ z² z³ θ θ = 45° each step
Repeated multiplication by a unit-modulus z walks evenly around the circle — De Moivre's theorem, visually.
Worked proof — De Moivre's theorem by induction

Base case (\(n=1\)): \((\cos\theta+i\sin\theta)^1 = \cos\theta+i\sin\theta\), true.

Inductive step: assume true for \(n=k\), i.e. \((\cos\theta+i\sin\theta)^k=\cos k\theta+i\sin k\theta\). Multiply both sides by \((\cos\theta+i\sin\theta)\):

\[(\cos\theta+i\sin\theta)^{k+1} = (\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta)\]

Expand, then use the angle-addition formulas \(\cos(A{+}B)=\cos A\cos B-\sin A\sin B\) and \(\sin(A{+}B)=\sin A\cos B+\cos A\sin B\):

\[=(\cos k\theta\cos\theta-\sin k\theta\sin\theta)+i(\sin k\theta\cos\theta+\cos k\theta\sin\theta)=\cos(k{+}1)\theta+i\sin(k{+}1)\theta\]

which is exactly the statement for \(n=k+1\). By induction, it holds for every positive integer \(n\).

6. The conjugate root theorem

The conjugate of \(z=a+bi\) is \(\bar z = a-bi\) — geometrically, \(z\) reflected across the real axis. Suppose \(p(z)\) is a polynomial with real coefficients and \(p(z)=0\). Taking the conjugate of both sides, every real coefficient survives untouched (conjugating a real number does nothing) — only \(z\) itself gets reflected — giving \(p(\bar z)=\overline{p(z)}=\bar 0=0\). So \(\bar z\) is a root too: for a real polynomial, non-real roots always come in mirror-image pairs across the real axis.

Diagram showing a complex root z above the real axis and its conjugate reflected below, connected by a dashed vertical line. z = a + bi (root) z̄ = a − bi (root) reflect
Non-real roots of a real polynomial mirror each other across the real axis.

7. The fundamental theorem of algebra

Put the two pictures together. The fundamental theorem of algebra says a degree-\(n\) polynomial has exactly \(n\) roots in the complex numbers, counted with multiplicity. For a real polynomial, non-real roots pair up by reflection (section 6); De Moivre's picture (section 5) says raising a unit-modulus number to successive powers walks it evenly around the circle. Put together: the \(n\) solutions of \(z^n=1\) sit at \(n\) evenly spaced points around the unit circle, each one \(360\degree/n\) further round than the last — a direct visual consequence of the rotation picture.

Unit circle with five evenly spaced points, the five fifth roots of unity, each 72 degrees apart. 1 ω ω² ω³ ω⁴ z⁵ = 1: 5 roots, 72° apart
The n roots of a real polynomial sit at n evenly-meaningful points — here, the 5th roots of unity.

Pictures help you remember. They don't replace practice.

These diagrams are here to make each result feel inevitable rather than arbitrary, and to give you something to picture when a question mentions Argand diagrams, De Moivre's theorem, or polynomial roots. They're not a substitute for the algebra — you still need to work through plenty of past-paper questions to be exam-ready.

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