Interpret conjugation translation rotation and complex division
Connect complex operations to transformations on the Argand plane, specifying their centre, direction and inverse.
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
- Carry out Cartesian and polar complex operations.
- Interpret directed translations, rotations and scale factors.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which operation translates every point by (3, −1)?
What you will learn
- Describe conjugation and addition or subtraction as geometric transformations.
- Calculate the image and inverse of a non-zero complex multiplication.
Conjugation reflects; addition translates
For z = x + iy, its conjugate x − iy reflects the point across the real axis. Points on the real axis stay fixed. Adding a fixed a = u + iv translates every point by vector (u,v), while subtracting a translates by (−u,−v). For z = 2 + i, its conjugate is 2 − i; adding 3 − i gives 5. These are different operations with different geometric effects.
A directed difference b − a represents the vector from a to b. Translation preserves distances and angles because it adds the same displacement to both endpoints. Conjugation also preserves lengths but reverses the orientation of a directed angle. Multiplication by −1 reflects through the origin, rather than across the real axis, so it is different from conjugation except at particular points.
A non-zero multiplier scales and rotates about the origin
Multiplication by q = rcis θ, r > 0, scales distances from the origin by r and rotates about the origin through θ. In particular multiplication by i is an anticlockwise rotation by π/2, because its modulus is one and its argument is π/2. The point 2 + i becomes −1 + 2i. Multiplication by 1 + i instead uses scale √2 and angle π/4, giving 1 + 3i.
Division by q is the inverse: scale 1/r and rotation −θ. To solve (1 + i)z = 4 + 2i, divide to obtain z = 3 − i. Applying the multiplier checks the inverse result. A rotation about another centre c needs a stated expression such as c + q(z − c); simply multiplying by q uses the origin. A zero multiplier collapses every point to zero and has no inverse.
Pause and explain
What does multiplication by i do?
Worked example
For z = 2 + i, calculate z*, iz and (1 + i)z and describe their transformations.
Show the worked solution
- Conjugation gives 2 − i, a reflection across the real axis.
- Multiplying by i gives −1 + 2i, a rotation by π/2 anticlockwise about the origin.
- Multiplying by 1 + i gives 1 + 3i, scale √2 and rotation π/4 about the origin.
Answer z* = 2 − i; iz = −1 + 2i; (1 + i)z = 1 + 3i.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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.
Translate z = −1 + 2i by adding 3 − i. Give its image and the inverse translation.
Give me a hint
Add the corresponding Cartesian parts and reverse the displacement to undo it.
Compare my reasoning
- The image is (−1 + 3) + (2 − 1)i = 2 + i.
- The displacement is (3, −1).
- Subtracting 3 − i from 2 + i returns −1 + 2i.
Image 2 + i; inverse subtracts 3 − i.
Look for these in your work
- Calculate the signed coordinate displacement.
- Check the inverse operation returns the starting point.
Apply multiplication by i to 3 − i. State the image, angle, centre and scale.
Give me a hint
Expand the Cartesian product and identify the multiplier’s modulus and argument.
Compare my reasoning
- i(3 − i) = 3i − i² = 1 + 3i.
- The multiplier i has modulus 1 and argument π/2.
- Thus the image is obtained by a π/2 anticlockwise rotation about the origin with scale 1.
Image 1 + 3i; rotation π/2 anticlockwise about the origin; scale 1.
Look for these in your work
- Use i² = −1 in the image calculation.
- Specify the rotation’s direction and centre.
A transformed point w = 4 + 2i was obtained by w = (1 + i)z. Find z with Cartesian working and describe the inverse geometry.
Give me a hint
Divide by 1 + i using its conjugate; the inverse scale and angle come from its polar form.
Compare my reasoning
- Use z = [(4 + 2i)(1 − i)]/2 = (6 − 2i)/2 = 3 − i.
- The multiplier has modulus √2 and argument π/4.
- The inverse scales by 1/√2 and rotates clockwise by π/4 about the origin; multiplication checks w.
z = 3 − i; inverse scale 1/√2 and rotation −π/4 about the origin.
Look for these in your work
- Verify the Cartesian inverse in the supplied mapping.
- State reciprocal scale and opposite rotation about the correct centre.
Common mistakes
- Conjugation is a real-axis reflection, not multiplication by −1.
- A multiplication’s rotation and enlargement use the origin unless a translated expression is supplied.
- An inverse needs a non-zero multiplier and uses reciprocal scale and opposite rotation.
What can you explain now?
Does multiplication by i change the modulus of a point?
Compare with the explanation
No: |i| = 1.
It rotates by π/2 about the origin and preserves all distances.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Calculate and describe the image of a point under a complex operation, then specify whether and how the operation can be reversed.
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.