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TASSpecialist MathematicsSyllabus dot point

How do we add, multiply, divide and conjugate complex numbers reliably in Cartesian form?

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

Adding, multiplying and dividing complex numbers in Cartesian form, conjugate properties, and realising denominators, with worked examples for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.78 min answer

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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 i2=βˆ’1i^2 = -1, so any power of ii collapses to one of 1,i,βˆ’1,βˆ’i1, i, -1, -i.

Powers of the imaginary unit

The powers of ii cycle with period four: i1=ii^1 = i, i2=βˆ’1i^2 = -1, i3=βˆ’ii^3 = -i, i4=1i^4 = 1, then repeat. To simplify a high power, divide the exponent by four and use the remainder. For example i23=i20β‹…i3=(i4)5β‹…i3=1β‹…(βˆ’i)=βˆ’ii^{23} = i^{20}\cdot i^3 = (i^4)^5 \cdot i^3 = 1 \cdot (-i) = -i.

Addition, subtraction and multiplication

For z=a+ibz = a + ib and w=c+idw = c + id:

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

Multiplication uses ordinary expansion, replacing i2i^2 with βˆ’1-1:

zw=(a+ib)(c+id)=ac+iad+ibc+i2bd=(acβˆ’bd)+i(ad+bc). 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

Conjugates obey clean rules that examiners love to test:

z+wβ€Ύ=zΛ‰+wΛ‰,zwβ€Ύ=zˉ wΛ‰,(zw)β€Ύ=zΛ‰wΛ‰. \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+zΛ‰=2Re⁑(z)z + \bar z = 2\operatorname{Re}(z) and zβˆ’zΛ‰=2iIm⁑(z)z - \bar z = 2i\operatorname{Im}(z). These let you extract the real or imaginary part without separating zz 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.

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.

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 practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 TASC3 marksIf z = 3 - 4i and w = 2 + i, determine the complex number v if 1/v = 1/z + 1/(w-bar).
Show worked answer β†’

First find each reciprocal by multiplying by the conjugate. 1/z = 1/(3 - 4i) = (3 + 4i)/((3 - 4i)(3 + 4i)) = (3 + 4i)/25.

The conjugate of w is w-bar = 2 - i, so 1/(w-bar) = 1/(2 - i) = (2 + i)/((2 - i)(2 + i)) = (2 + i)/5 = (10 + 5i)/25.

Add: 1/v = (3 + 4i)/25 + (10 + 5i)/25 = (13 + 9i)/25.

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

2023 TASC4 marksEvaluate 1/(2 + i)^2 - 1/(2 - i)^2. Give your answer in polar form.
Show worked answer β†’

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: 1/(3 + 4i) - 1/(3 - 4i) = [(3 - 4i) - (3 + 4i)] / [(3 + 4i)(3 - 4i)] = (-8i)/(9 + 16) = -8i/25.

This is a purely imaginary number with modulus 8/25 and argument -pi/2. In polar form it is (8/25)(cos(-pi/2) + i sin(-pi/2)), or equivalently (8/25) e^(-i pi/2). Markers reward using conjugate pairs to clear the denominator and correctly reporting both modulus and argument.