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Module 12 · L03 of 16 ~40 min ⚡ +90 XP available

Conjugates and Division

Complex division has no direct meaning — until you meet the conjugate. The conjugate $\bar{z} = a - bi$ is the algebraic partner that flips a complex number across the real axis, turns any complex denominator into a real number, and makes division as routine as multiplication. By the end of this lesson you'll wield conjugate properties fluently and divide any pair of complex numbers in $a + bi$ form.

Today's hook — Without using a calculator, try to write $\dfrac{1}{1 + i}$ in the form $a + bi$. Where do you get stuck? What single trick should turn the denominator into a real number? Compare your strategy after card 05.

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Recall — your gut answer first

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For $z = 3 + 4i$, write down what you think $\bar{z}$ should be. Then compute $z \cdot \bar{z}$ (expand fully). Before checking — is the answer real? What does it equal in terms of $|z|$?

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The two moves for complex conjugates

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Every problem in this lesson rewards two habits: flip the imaginary sign to form the conjugate, then multiply top and bottom by the conjugate when a complex number sits in a denominator. Together these turn division into a routine real-arithmetic problem.

The conjugate–multiply–simplify reading: (1) identify the complex denominator $c + di$, (2) multiply numerator and denominator by its conjugate $c - di$, (3) the new denominator is $c^2 + d^2$ — a positive real — so the result is easy to split into $a + bi$.

Conjugate: $\overline{a+bi} = a - bi$ · $z\bar{z} = a^2 + b^2 = |z|^2$ · $\dfrac{z_1}{z_2} = \dfrac{z_1 \bar{z_2}}{z_2 \bar{z_2}}$

$\dfrac{1}{c + di} = \dfrac{c - di}{c^2 + d^2}$

Conjugate flips the sign of $i$

If $z = a + bi$ then $\bar{z} = a - bi$. Geometrically this is a reflection across the real axis. Real numbers are their own conjugates; purely imaginary numbers flip sign.

$z\bar{z}$ is always real and non-negative

$z\bar{z} = a^2 + b^2 = |z|^2$. This identity is the engine of complex division — it converts a complex denominator into a single positive real.

Conjugation distributes

$\overline{z + w} = \bar{z} + \bar{w}$ and $\overline{zw} = \bar{z}\,\bar{w}$. So conjugation commutes with addition, subtraction, multiplication and (non-zero) division.

What you'll master

Know

Key facts

$\overline{a + bi} = a - bi$ (definition)
$z + \bar{z} = 2\,\text{Re}(z)$ and $z - \bar{z} = 2i\,\text{Im}(z)$
$z\bar{z} = a^2 + b^2 = |z|^2$ (a non-negative real)
$\overline{z + w} = \bar{z} + \bar{w}$ and $\overline{zw} = \bar{z}\,\bar{w}$

Understand

Concepts

Why conjugation is reflection across the real axis
Why $z\bar{z}$ is always real — and why this makes complex division possible
Why $z$ is real iff $z = \bar{z}$, and purely imaginary iff $z = -\bar{z}$

Can do

Skills

Form the conjugate of any complex expression
Divide complex numbers by rationalising with the conjugate
Simplify any expression of the form $\dfrac{a + bi}{c + di}$ into $x + yi$

Key terms

Complex conjugate ($\bar{z}$)For $z = a + bi$, the conjugate is $\bar{z} = a - bi$. Geometrically: reflection of $z$ in the real axis on the Argand plane.

$\text{Re}(z)$, $\text{Im}(z)$Real and imaginary parts of $z = a + bi$: $\text{Re}(z) = a$, $\text{Im}(z) = b$. Both are real numbers; $\text{Im}(z)$ does NOT include the $i$.

Modulus squared identity$z\bar{z} = (a+bi)(a-bi) = a^2 + b^2 = |z|^2$. Always a non-negative real; zero only when $z = 0$.

Rationalising the denominatorTechnique for complex division: multiply numerator and denominator by the conjugate of the denominator so the new denominator becomes a real number.

Conjugate distributivity$\overline{z + w} = \bar{z} + \bar{w}$, $\overline{zw} = \bar{z}\,\bar{w}$, and $\overline{(z/w)} = \bar{z}/\bar{w}$ (provided $w \ne 0$). Conjugation behaves like a field automorphism on $\mathbb{C}$.

Reality test$z \in \mathbb{R}$ iff $z = \bar{z}$; $z$ is purely imaginary iff $z = -\bar{z}$ (and $z \ne 0$). Used to extract real/imaginary parts cleanly.

MEX-N1NESA outcome (Complex Numbers I): performs operations with complex numbers including addition, subtraction, multiplication, division and conjugation in Cartesian form.

The conjugate and its properties

core concept

For $z = a + bi$ with $a, b \in \mathbb{R}$, the complex conjugate is $\bar{z} = a - bi$. Three identities follow immediately from this definition:

Sum: $z + \bar{z} = (a + bi) + (a - bi) = 2a = 2\,\text{Re}(z)$.
Difference: $z - \bar{z} = (a + bi) - (a - bi) = 2bi = 2i\,\text{Im}(z)$.
Product: $z \bar{z} = (a+bi)(a-bi) = a^2 - (bi)^2 = a^2 + b^2 = |z|^2$.

The product identity is the workhorse of this lesson. It tells us that multiplying any complex number by its conjugate eliminates the imaginary part entirely, producing a non-negative real number.

Distributivity of conjugation. Let $z = a + bi$ and $w = c + di$. Then $z + w = (a+c) + (b+d)i$ and $\overline{z + w} = (a+c) - (b+d)i = (a - bi) + (c - di) = \bar{z} + \bar{w}$. A similar (slightly longer) calculation shows $\overline{zw} = \bar{z}\,\bar{w}$.

Connecting to division. To compute $\dfrac{z}{w}$ when $w$ is complex, multiply top and bottom by $\bar{w}$. The new denominator is $w\bar{w} = |w|^2$ — a positive real. The result splits cleanly into $a + bi$ form.

$\overline{a + bi} = a - bi$ (flip the imaginary sign) · $z + \bar{z} = 2\,\text{Re}(z)$; $z - \bar{z} = 2i\,\text{Im}(z)$; $z\bar{z} = |z|^2$ · $\overline{z + w} = \bar{z} + \bar{w}$ and $\overline{zw} = \bar{z}\,\bar{w}$ · Complex division: multiply top and bottom by the conjugate of the denominator

Pause — copy $\overline{a+bi} = a-bi$, the identities $z+\bar{z} = 2\operatorname{Re}(z)$, $z\bar{z} = |z|^2$, and $\overline{zw} = \bar{z}\bar{w}$ into your book.

Quick check: Let $z = 5 - 2i$. Which expression equals $z\bar{z}$?

Division by rationalising the denominator

core concept

We just saw that $\bar{z} = a - bi$ satisfies $z\bar{z} = |z|^2$, $\overline{z+w} = \bar{z}+\bar{w}$, and $\overline{zw} = \bar{z}\bar{w}$. That raises a question: how do we carry out division $z_1/z_2$ and express the result in standard $a+bi$ form? This card answers it → multiply numerator and denominator by $\bar{z}_2$, making the denominator the real number $|z_2|^2 = c^2+d^2$.

To divide $\dfrac{z_1}{z_2}$ where $z_2 = c + di \ne 0$, multiply numerator and denominator by $\bar{z_2} = c - di$:

$$\frac{z_1}{z_2} = \frac{z_1}{z_2} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{z_1 \bar{z_2}}{|z_2|^2}$$

Because $|z_2|^2 = c^2 + d^2$ is a positive real, the result is now a complex number divided by a real number — separate the real and imaginary parts and you have $a + bi$ form.

Why this works. The trick is the same one used to rationalise surd denominators ($\dfrac{1}{\sqrt{2} - 1} \cdot \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1}$). For complex numbers, the "irrationality" we are removing is the imaginary part $i$, and the conjugate is the algebraic partner that cancels it.

Common mistake. Do not write the answer as a single fraction unless asked. The standard form is $a + bi$ — i.e., $\dfrac{1 + 3i}{1 + i} = 2 + i$, not $\dfrac{4 + 2i}{2}$. Always finish the division.

$\dfrac{z_1}{z_2} = \dfrac{z_1 \bar{z_2}}{|z_2|^2}$ — multiply by the conjugate of the denominator · New denominator $|z_2|^2 = c^2 + d^2$ is a positive real · Always finish: split the fraction into $a + bi$ form · Mirror of surd rationalisation: $\dfrac{1}{c + di} \cdot \dfrac{c - di}{c - di}$

Pause — copy the division formula $z_1/z_2 = z_1\bar{z}_2/|z_2|^2$, the step of multiplying by the conjugate of the denominator, and the instruction to split the result into $a+bi$ form into your book.

Did you get this? True or false: for any non-zero complex number $z$, the product $z\bar{z}$ is always a positive real number.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BASIC DIVISION

Express $\dfrac{1}{1 + i}$ in the form $a + bi$ where $a, b \in \mathbb{R}$.

Identify the denominator: $1 + i$. Its conjugate is $1 - i$. Multiply numerator and denominator by $1 - i$: $\dfrac{1}{1 + i} \cdot \dfrac{1 - i}{1 - i} = \dfrac{1 - i}{(1 + i)(1 - i)}$.

The conjugate of $c + di$ is $c - di$. Multiplying by $\dfrac{\bar{w}}{\bar{w}}$ does not change the value (it's multiplying by 1), but it transforms the denominator into a real number.

PROBLEM 2 · DIVISION WITH BOTH PARTS NON-ZERO

Express $\dfrac{3 + 2i}{1 - i}$ in the form $a + bi$.

Denominator $1 - i$ has conjugate $1 + i$. Multiply top and bottom: $\dfrac{3 + 2i}{1 - i} \cdot \dfrac{1 + i}{1 + i} = \dfrac{(3 + 2i)(1 + i)}{(1 - i)(1 + i)}$.

Multiplying by $\dfrac{1+i}{1+i}$ preserves the value (it's 1) but rationalises the denominator. The conjugate of $c - di$ is $c + di$ — sign-flip the imaginary part.

PROBLEM 3 · USING CONJUGATE PROPERTIES

Let $z = 2 + 3i$ and $w = 1 - 4i$. Verify directly that $\overline{z + w} = \bar{z} + \bar{w}$, then compute $\dfrac{\bar{z}}{w}$ in $a + bi$ form.

$z + w = (2 + 3i) + (1 - 4i) = 3 - i$, so $\overline{z + w} = 3 + i$. Separately, $\bar{z} = 2 - 3i$ and $\bar{w} = 1 + 4i$, so $\bar{z} + \bar{w} = 3 + i$. ✓ Equal.

Conjugation distributes over addition. Compute both sides and check they agree — this confirms the identity $\overline{z + w} = \bar{z} + \bar{w}$.

Fill the gap: For $z = a + bi$, the conjugate is $\bar{z} = a$ $bi$, and the product $z\bar{z} = $ which equals $|z|^2$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting to multiply the numerator

To divide, you must multiply BOTH numerator and denominator by the conjugate. Multiplying only the denominator changes the value of the fraction. The conjugate trick relies on $\dfrac{\bar{w}}{\bar{w}} = 1$.

Trap 02

Sign error when forming the conjugate

For $z = 3 - 4i$, $\bar{z} = 3 + 4i$ (flip the sign of the imaginary part). Students often write $\bar{z} = -3 - 4i$ or $-3 + 4i$. Only the imaginary part changes sign — the real part is unchanged.

Trap 03

Leaving the answer as a single fraction

The standard form is $a + bi$. Writing $\dfrac{4 + 2i}{2}$ is not finished — simplify to $2 + i$. NESA markers want answers in the canonical Cartesian form unless asked otherwise.

Did you get this? True or false: the conjugate of $-2 + 5i$ is $2 - 5i$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Write the conjugate of each: (a) $4 + 7i$, (b) $-3 - i$, (c) $5$, (d) $-2i$.

For $z = 2 - 5i$, compute $z + \bar{z}$, $z - \bar{z}$, and $z\bar{z}$. Confirm $z\bar{z} = |z|^2$.

Express $\dfrac{2}{3 - i}$ in $a + bi$ form.

Express $\dfrac{4 + 3i}{2 + i}$ in $a + bi$ form.

Let $z = 1 + 2i$ and $w = 3 + i$. Verify $\overline{zw} = \bar{z}\,\bar{w}$ by computing both sides.

Odd one out: Three of these expressions equal a real number for every complex $z$. Which one does NOT?

Revisit your thinking

At the start you tried $\dfrac{1}{1 + i}$ and (probably) hit a wall — direct division of complex numbers has no obvious meaning.

The trick: multiply top and bottom by the conjugate $1 - i$. The denominator becomes $(1+i)(1-i) = 2$ — a real number — and the rest is real arithmetic: $\dfrac{1 - i}{2} = \dfrac{1}{2} - \dfrac{1}{2}i$. The conjugate is the key that unlocks division. Every complex-fraction question on the HSC paper reduces to this single move. Memorise the identity $z\bar{z} = a^2 + b^2 = |z|^2$ and division becomes routine.

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Multiple choice

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Short answer

ApplyBand 32 marks

Q1. Express $\dfrac{5}{2 + i}$ in the form $a + bi$. (2 marks)

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ApplyBand 43 marks

Q2. Let $z = 3 + 2i$ and $w = 1 + 4i$. Compute $\dfrac{z}{w}$ in $a + bi$ form. (3 marks)

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AnalyseBand 53 marks

Q3. Prove that for any complex numbers $z$ and $w$ with $w \ne 0$, $\overline{\left(\dfrac{z}{w}\right)} = \dfrac{\bar{z}}{\bar{w}}$. You may assume $\overline{zw} = \bar{z}\,\bar{w}$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. (a) $4 - 7i$. (b) $-3 + i$. (c) $5$ (any real number is self-conjugate). (d) $2i$.

2. $z + \bar{z} = (2 - 5i) + (2 + 5i) = 4 = 2 \cdot \text{Re}(z)$. $z - \bar{z} = -10i = 2i \cdot \text{Im}(z)$. $z\bar{z} = (2)^2 + (-5)^2 = 29 = |z|^2$ ✓.

3. $\dfrac{2}{3 - i} \cdot \dfrac{3 + i}{3 + i} = \dfrac{2(3 + i)}{9 + 1} = \dfrac{6 + 2i}{10} = \dfrac{3}{5} + \dfrac{1}{5}i$.

4. $\dfrac{4 + 3i}{2 + i} \cdot \dfrac{2 - i}{2 - i} = \dfrac{(4 + 3i)(2 - i)}{5} = \dfrac{8 - 4i + 6i - 3i^2}{5} = \dfrac{11 + 2i}{5} = \dfrac{11}{5} + \dfrac{2}{5}i$.

5. $zw = (1 + 2i)(3 + i) = 3 + i + 6i + 2i^2 = 1 + 7i$, so $\overline{zw} = 1 - 7i$. $\bar{z}\,\bar{w} = (1 - 2i)(3 - i) = 3 - i - 6i + 2i^2 = 1 - 7i$. Equal ✓.

Q1 (2 marks): $\dfrac{5}{2 + i} \cdot \dfrac{2 - i}{2 - i} = \dfrac{5(2 - i)}{4 + 1} = \dfrac{10 - 5i}{5}$ [1] $= 2 - i$ [1].

Q2 (3 marks): Multiply by $\dfrac{1 - 4i}{1 - 4i}$ [1]. Numerator: $(3 + 2i)(1 - 4i) = 3 - 12i + 2i - 8i^2 = 11 - 10i$. Denominator: $1 + 16 = 17$ [1]. So $\dfrac{z}{w} = \dfrac{11}{17} - \dfrac{10}{17}i$ [1].

Q3 (3 marks): Let $q = \dfrac{z}{w}$, so $z = qw$ [1]. Conjugating: $\bar{z} = \overline{qw} = \bar{q}\,\bar{w}$ (given) [1]. Since $\bar{w} \ne 0$, divide: $\bar{q} = \dfrac{\bar{z}}{\bar{w}}$, i.e., $\overline{\left(\dfrac{z}{w}\right)} = \dfrac{\bar{z}}{\bar{w}}$ [1].

Boss battle · The Conjugate Closer

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Five timed questions on conjugates, $z\bar{z}$ identities and complex division. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering quick complex-arithmetic questions. Lighter alternative to the boss.

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