Calculate Cartesian products quotients and conjugates
Add and subtract components, expand products with i² = −1 and divide by multiplying by the denominator’s conjugate.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.9. Paper 1 algebra and radian trigonometry are assumed. Use Cartesian working for products, quotients and the two square roots of a non-zero complex number. Arguments usually use −π < θ ≤ π; an alternative interval 0 ≤ θ < 2π is acceptable unless specified. The argument of zero is undefined. General nth-root methods are beyond this mapped sequence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Collect signed real and imaginary terms.
- Expand binomials and simplify fractions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the conjugate of 2 − 3i?
What you will learn
- Carry out Cartesian addition, subtraction and multiplication with full working.
- Use a conjugate to divide by a non-zero complex number.
Addition and multiplication use different rules
For z = x + iy and w = a + ib, add or subtract corresponding parts: z + w = (x + a) + i(y + b). For multiplication expand every term and use i² = −1: zw = (xa − yb) + i(xb + ya). With z = 3 + 2i and w = 1 − i, their sum is 4 + i, difference is 2 + 3i and product is 5 − i. Multiplying imaginary coefficients without the negative i² contribution changes the real part incorrectly.
The conjugate z* = x − iy reflects the sign of the imaginary part. The product zz* = x² + y² is real and non-negative, and vanishes only at z = 0. For 3 + 2i the conjugate is 3 − 2i and their product is 13. Conjugation preserves a real number and reverses a purely imaginary number; it does not mean changing the signs of both parts.
A conjugate makes a complex denominator real
To calculate z/w, w ≠ 0, multiply numerator and denominator by w*. The denominator becomes a² + b², so the quotient is [(xa + yb) + i(ya − xb)]/(a² + b²). Show the expansion and simplification when Cartesian working is required. For (3 + 2i)/(1 − i), multiply by 1 + i to obtain (1 + 5i)/2.
A complex equation such as (1 + 2i)z = 3 − i can be solved by this same division because the multiplier is non-zero. Then check the proposed quotient by multiplying it by the original denominator. Dividing by zero is undefined in the complex system just as it is in real arithmetic; the conjugate method cannot make a zero denominator valid.
Pause and explain
Which factor makes the denominator 2 − i real?
Worked example
Calculate (3 + 2i)/(1 − i), showing Cartesian working.
Show the worked solution
- Multiply top and bottom by 1 + i.
- Expand the numerator: 3 + 3i + 2i + 2i² = 1 + 5i; the denominator is 1 − i² = 2.
- Thus the quotient is (1 + 5i)/2; multiplication by 1 − i returns 3 + 2i.
Answer (1 + 5i)/2.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate (2 − 3i)(−1 + 4i), showing the expansion.
Give me a hint
The product of the two imaginary terms uses i² = −1.
Compare my reasoning
- Expand to −2 + 8i + 3i − 12i².
- Replace i² by −1, so the real terms total 10.
- Collect the imaginary terms to obtain 10 + 11i.
10 + 11i.
Look for these in your work
- Show all four terms in the Cartesian product.
- Use the correct sign for the i² term.
Calculate (4 + i)/(2 − i) in Cartesian form with full working.
Give me a hint
Multiply both parts of the fraction by the conjugate 2 + i.
Compare my reasoning
- Use [(4 + i)(2 + i)]/[(2 − i)(2 + i)].
- The numerator is 8 + 6i + i² = 7 + 6i, and the denominator is 5.
- Hence z = (7 + 6i)/5; multiplying by 2 − i recovers 4 + i.
(7 + 6i)/5.
Look for these in your work
- Use the denominator’s conjugate on top and bottom.
- Simplify and verify the original quotient.
Solve (1 + 2i)z = 3 − i in Cartesian form and verify your answer.
Give me a hint
Divide by the non-zero multiplier using its conjugate.
Compare my reasoning
- Multiply the quotient (3 − i)/(1 + 2i) by (1 − 2i)/(1 − 2i).
- The numerator is 1 − 7i and the denominator is 5, giving z = (1 − 7i)/5.
- Multiplication by 1 + 2i gives (15 − 5i)/5 = 3 − i.
z = (1 − 7i)/5.
Look for these in your work
- Solve by a valid non-zero division.
- Verify the result in the original complex equation.
Common mistakes
- Expand all four product terms and replace i² by −1.
- Conjugation changes only the imaginary sign.
- Multiply numerator and denominator by the same conjugate and exclude a zero denominator.
What can you explain now?
What is (2 + i)(2 − i)?
Compare with the explanation
5.
The middle terms cancel and −i² contributes +1 to the real product.
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Make it stick
Expand a Cartesian product, form a conjugate and divide with a real denominator, showing each algebraic step.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.