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Proof and Complex Numbers

Specialist Mathematics
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Proof and Complex Numbers

Specialist Mathematics
01 May 2026

Proof and Complex Numbers

This topic introduces the formal language of mathematics and the extension of the number system to include complex numbers.

The Role of Proof

Term Definition
Axiom A statement accepted without proof
Definition Precise meaning assigned to a term
Theorem A statement proved from axioms and prior results
Lemma A helper result used to prove a larger theorem
Corollary A result following directly from a theorem

Core Proof Techniques

Direct proof: Assume hypothesis, apply logic, reach conclusion.

Contrapositive: To prove $P \Rightarrow Q$, prove the equivalent $\neg Q \Rightarrow \neg P$.

Contradiction: Assume $\neg P$; derive a known falsehood.

Mathematical induction:
1. Base case: verify $P(1)$.
2. Inductive step: assume $P(k)$, prove $P(k+1)$.
3. Conclude $P(n)$ holds for all $n \geq 1$.

Induction example: Prove $\displaystyle\sum_{r=1}^{n} r = \dfrac{n(n+1)}{2}$.

Base case $n=1$: LHS $= 1$, RHS $= \frac{1 \cdot 2}{2} = 1$. True.

Inductive step: Assume $\displaystyle\sum_{r=1}^{k} r = \frac{k(k+1)}{2}$. Then:
$$\sum_{r=1}^{k+1} r = \frac{k(k+1)}{2} + (k+1) = \frac{(k+1)(k+2)}{2}.$$
This is the formula at $n = k+1$. $\square$

The Complex Number Field

$\mathbb{C} = {a + bi : a, b \in \mathbb{R}}$ where $i^2 = -1$.

  • Conjugate: $\bar{z} = a - bi$
  • Modulus: $|z| = \sqrt{a^2 + b^2}$
  • Argument: $\arg(z) = \theta$ with $\cos\theta = a/|z|$, $\sin\theta = b/|z|$

Division: $\dfrac{z_1}{z_2} = \dfrac{z_1 \bar{z}_2}{|z_2|^2}$.

Example: $z = 3+4i$: $|z| = 5$, $\arg(z) = \arctan(4/3)$, $\dfrac{1}{z} = \dfrac{3-4i}{25}$.

KEY TAKEAWAY: Proof and complex numbers are the twin foundations of Specialist Mathematics. Master the four proof methods and the arithmetic of $\mathbb{C}$ before attempting any other topic.

EXAM TIP: In induction proofs, always label the base case and inductive step. VCAA awards marks for structure.

COMMON MISTAKE: Confusing the contrapositive ($\neg Q \Rightarrow \neg P$, logically equivalent) with the converse ($Q \Rightarrow P$, not equivalent).

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