Complex arithmetic and conjugates - TCE Mathematics Specialised (Tasmania)

Syllabus dot point

“Perform complex arithmetic in Cartesian form and use conjugate properties to divide and simplify.”

TCESpecialist MathematicsUnit 38 min readUpdated 2 Jun 2026

Quick answer

Add and subtract complex numbers component by component, multiply using the distributive law with $i^2 = -1$ , and divide by multiplying top and bottom by the conjugate of the denominator. The conjugate $\bar z = x - iy$ is the workhorse, because $z\bar z = x^2 + y^2$ is always a non-negative real number.

What this dot point is asking

This dot point is the foundation that every later complex number skill stands on. Before you can use polar form, De Moivre or factorisation, you must be able to do arithmetic in Cartesian form without slips. The single idea that drives all of it is that $i^2 = -1$ , so any power of $i$ collapses to one of $1, i, -1, -i$ .

Powers of the imaginary unit

The powers of $i$ cycle with period four: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ , then repeat. To simplify a high power, divide the exponent by four and use the remainder. For example $i^{23} = i^{20}\cdot i^3 = (i^4)^5 \cdot i^3 = 1 \cdot (-i) = -i$ .

Addition, subtraction and multiplication

For $z = a + ib$ and $w = c + id$ :

z + w = (a + c) + i(b + d), \qquad z - w = (a - c) + i(b - d).

Multiplication uses ordinary expansion, replacing $i^2$ with $-1$ :

zw = (a + ib)(c + id) = ac + iad + ibc + i^2 bd = (ac - bd) + i(ad + bc).

You do not need to memorise that final formula. It is faster and safer to expand the brackets each time and tidy up.

The conjugate

Complex conjugate

The conjugate of $z = x + iy$ is $\bar z = x - iy$ . Geometrically it is the reflection of $z$ in the real axis. Its defining usefulness is that

z\bar z = (x + iy)(x - iy) = x^2 + y^2 = |z|^2,

which is always a real number greater than or equal to zero.

Conjugates obey clean rules that examiners love to test:

\overline{z + w} = \bar z + \bar w, \qquad \overline{zw} = \bar z\,\bar w, \qquad \overline{\left(\tfrac{z}{w}\right)} = \frac{\bar z}{\bar w}.

Two further identities are worth knowing: $z + \bar z = 2\operatorname{Re}(z)$ and $z - \bar z = 2i\operatorname{Im}(z)$ . These let you extract the real or imaginary part without separating $z$ into pieces by hand.

Division by realising the denominator

To divide, multiply numerator and denominator by the conjugate of the denominator. This turns the denominator into a real number, after which you split into real and imaginary parts.

Dividing complex numbers

Express $\dfrac{3 + 2i}{1 - 4i}$ in the form $x + iy$ .

Step 1: Identify the conjugate of the denominator

The denominator is $1 - 4i$ , so its conjugate is $1 + 4i$ . Multiplying both numerator and denominator by this conjugate is the key move: it turns the denominator into a real number via the identity $z\bar{z} = x^2 + y^2$ , which contains no $i$ at all.

\frac{3 + 2i}{1 - 4i}\cdot\frac{1 + 4i}{1 + 4i} = \frac{(3 + 2i)(1 + 4i)}{1^2 + 4^2}.

Step 2: Expand the numerator

We now expand the product in the numerator using the distributive law, remembering that $i^2 = -1$ so any $i^2$ term contributes a real number with the opposite sign.

(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i.

Step 3: Evaluate the denominator and write in $x + iy$ form

The denominator is the sum of squares $1^2 + 4^2 = 17$ , a real number, so we can now split into real and imaginary parts straightforwardly.

\frac{3 + 2i}{1 - 4i} = \frac{-5 + 14i}{17} = -\frac{5}{17} + \frac{14}{17}\,i.

Final answer: $-\dfrac{5}{17} + \dfrac{14}{17}\,i$ .

Equating real and imaginary parts

Two complex numbers are equal exactly when their real parts match and their imaginary parts match. This turns one complex equation into two real equations, a technique used constantly when solving for unknowns.

Solving by equating parts

Find real numbers $a$ and $b$ such that $(a + ib)(2 - i) = 5 + 5i$ .

Step 1: Expand the left side and collect real and imaginary parts

Because two complex numbers are equal if and only if their real parts match and their imaginary parts match separately, we first need the left side in the form $\text{real} + i\cdot\text{imaginary}$ . Expanding using the distributive law and replacing $i^2$ with $-1$ gives:

(a + ib)(2 - i) = 2a - ai + 2bi - bi^2 = (2a + b) + i(2b - a).

Step 2: Set up a system of two real equations

Now that both sides are in Cartesian form, we equate the real part of the left side to the real part of the right side, and likewise for the imaginary parts. This converts one complex equation into two ordinary simultaneous equations in the unknowns $a$ and $b$ .

2a + b = 5, \qquad 2b - a = 5.

Step 3: Solve the simultaneous equations

From the first equation, $b = 5 - 2a$ . Substituting into the second eliminates $b$ , leaving a single equation in $a$ .

2(5 - 2a) - a = 5 \implies 10 - 5a = 5 \implies a = 1, \quad b = 3.

Check: $(1 + 3i)(2 - i) = 2 - i + 6i + 3 = 5 + 5i$ . Correct.

Final answer: $a = 1$ , $b = 3$ .

Conjugate of a real-coefficient expression

If a polynomial has only real coefficients, then taking the conjugate of the whole equation simply replaces every $z$ by $\bar z$ . This is why non-real roots of real polynomials always come in conjugate pairs, a fact you will use heavily in factorisation.

Forgetting that

i^2 = -1

during expansion

The most common arithmetic error is treating $i^2$ as if it were $+1$ , or leaving an $i^2$ unsimplified. Every time a product of two imaginary parts appears, it contributes a real term with the opposite sign. Always sweep through your expansion and replace each $i^2$ with $-1$ before collecting like terms, then check that your final answer has a clearly separated real and imaginary part.

Why this matters

Fluent Cartesian arithmetic is what lets you switch to polar form with confidence and check polar answers by converting back. Practise division until realising the denominator is automatic, because it appears inside almost every harder complex number question.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks

z = 3 - 4i

and

w = 2 + i

, determine the complex number

v

where

\dfrac{1}{v} = \dfrac{1}{z} + \dfrac{1}{\bar{w}}

Show worked answer →

First find each reciprocal by multiplying by the conjugate. $\dfrac{1}{z} = \dfrac{1}{3 - 4i} = \dfrac{3 + 4i}{(3 - 4i)(3 + 4i)} = \dfrac{3 + 4i}{25}$ .

The conjugate of $w$ is $\bar{w} = 2 - i$ , so $\dfrac{1}{\bar{w}} = \dfrac{1}{2 - i} = \dfrac{2 + i}{(2 - i)(2 + i)} = \dfrac{2 + i}{5} = \dfrac{10 + 5i}{25}$ .

Add: $\dfrac{1}{v} = \dfrac{3 + 4i}{25} + \dfrac{10 + 5i}{25} = \dfrac{13 + 9i}{25}$ .

Therefore $v = \dfrac{25}{13 + 9i} = \dfrac{25(13 - 9i)}{(13 + 9i)(13 - 9i)} = \dfrac{25(13 - 9i)}{169 + 81} = \dfrac{25(13 - 9i)}{250} = \dfrac{13 - 9i}{10}$ . So $v = 1.3 - 0.9i$ . Markers reward realising each denominator with its conjugate and a correct final division.

Practice4 marks

Evaluate

\dfrac{1}{(2 + i)^2} - \dfrac{1}{(2 - i)^2}

. Give your answer in polar form.

Show worked answer →

First square the denominators: $(2 + i)^2 = 4 + 4i + i^2 = 3 + 4i$ , and $(2 - i)^2 = 3 - 4i$ (its conjugate).

Combine over a common denominator: $\dfrac{1}{3 + 4i} - \dfrac{1}{3 - 4i} = \dfrac{(3 - 4i) - (3 + 4i)}{(3 + 4i)(3 - 4i)} = \dfrac{-8i}{9 + 16} = -\dfrac{8i}{25}$ .

This is a purely imaginary number with modulus $\dfrac{8}{25}$ and argument $-\dfrac{\pi}{2}$ . In polar form it is $\dfrac{8}{25}\operatorname{cis}\!\left(-\dfrac{\pi}{2}\right)$ , or equivalently $\dfrac{8}{25}e^{-i\pi/2}$ . Markers reward using conjugate pairs to clear the denominator and correctly reporting both modulus and argument.

Practise this

Sources & how we know this

Mathematics Specialised Level 4 (MTS415118) course document (2018 onwards) — TASC

What this dot point is asking

Powers of the imaginary unit

Addition, subtraction and multiplication

The conjugate

Division by realising the denominator

Step 1: Identify the conjugate of the denominator

Step 2: Expand the numerator

Step 3: Evaluate the denominator and write in x+iyx + iyx+iy form

Equating real and imaginary parts

Step 1: Expand the left side and collect real and imaginary parts

Step 2: Set up a system of two real equations

Step 3: Solve the simultaneous equations

Why this matters

Exam-style questions

Practise this

Sources & how we know this

Step 3: Evaluate the denominator and write in $x + iy$ form