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.
+5 XP warm-up For $z = 3 + 4i$, write down what you think $\bar{z}$ should be.
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 $|z|$?
+5 XP to read Every problem in this lesson rewards two habits: flip the imaginary sign to form the conjugate, then multiply...
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: $\overline{a+bi} = a - bi$ · $z\bar{z} = a^2 + b^2 = |z|^2$ · $\dfrac{z_1}{z_2} = \dfrac{z_1 \bar{z_2}}{z_2 \bar{z_2}}$
For $z = a + bi$ with $a, b \in \mathbb{R}$, the complex conjugate is $\bar{z} = a - bi$.
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:
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 $\overline{zw} = \bar{z}\,\bar{w}$.
$\overline{a + bi} = a - bi$ (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 $z\bar{z}$?
We just saw that $\bar{z} = a - bi$ satisfies $z\bar{z} = |z|^2$, $\overline{z+w} = \bar{z}+\bar{w}$, and $\overline{zw} = \bar{z}\bar{w}$.
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 $\bar{z_2} = c - di$:
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.
$\dfrac{z_1}{z_2} = \dfrac{z_1 \bar{z_2}}{|z_2|^2}$, 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
Worked examples · 3 in a row, reveal as you go
Express $\dfrac{1}{1 + i}$ in the form $a + bi$ where $a, b \in \mathbb{R}$.
Express $\dfrac{3 + 2i}{1 - i}$ in the form $a + bi$.
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.
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
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the conjugate of $-2 + 5i$ is $2 - 5i$.
Activities · practice with the ideas
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?
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.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Express $\dfrac{5}{2 + i}$ in the form $a + bi$. (2 marks)
Q2. Let $z = 3 + 2i$ and $w = 1 + 4i$. Compute $\dfrac{z}{w}$ in $a + bi$ form. (3 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)
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].
Sit the module quiz, then close the lesson off.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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