TASC TCE 2024-style practice: If z = 3 - 4i and w = 2 + i, determine the complex number v where 1v = 1z + 1w.

First find each reciprocal by multiplying by the conjugate. 1z = 13 - 4i = 3 + 4i(3 - 4i)(3 + 4i) = 3 + 4i25. The conjugate of w is w = 2 - i, so 1w = 12 - i = 2 + i(2 - i)(2 + i) = 2 + i5 = 10 + 5i25. Add: 1v = 3 + 4i25 + 10 + 5i25 = 13 + 9i25. Therefore v = 2513 + 9i = 25(13 - 9i)(13 + 9i)(13 - 9i) = 25(13 - 9i)169 + 81 = 25(13 - 9i)250 = 13 - 9i10. So v = 1.3 - 0.9i. Markers reward realising each denominator with its conjugate and a correct final division.

TASC TCE 2023-style practice: Evaluate 1(2 + i)^2 - 1(2 - i)^2. Give your answer in polar form.

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: 13 + 4i - 13 - 4i = (3 - 4i) - (3 + 4i)(3 + 4i)(3 - 4i) = -8i9 + 16 = -8i25. This is a purely imaginary number with modulus 825 and argument -π2. In polar form it is 825cis\!(-π2), or equivalently 825e^-iπ/2. Markers reward using conjugate pairs to clear the denominator and correctly reporting both modulus and argument.

§-Syllabus dot point

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 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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 $i^2 = -1$ , 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

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

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.

TCE 20243 marksIf

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.

TCE 20234 marksEvaluate

\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.