Understand complex numbers and equate real and imaginary parts
Use i² = −1, identify the two real components of a complex number and solve equality conditions without treating i as an unknown.
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
- Expand expressions and collect like terms.
- Solve two simultaneous linear equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is i²?
What you will learn
- Identify the real and imaginary parts of a Cartesian complex number.
- Equate corresponding parts to determine real parameters.
A complex number has two real components
The imaginary unit i satisfies i² = −1. A complex number z = x + iy has real components x and y; Re z = x and Im z = y. The imaginary part is the real coefficient y, not iy. Thus Re(4 − 3i) = 4 and Im(4 − 3i) = −3. A real number is included by taking y = 0, so the real numbers form part of the complex system. There is no real solution to u² = −1, but the complex solutions are i and −i.
Use i² = −1 whenever two imaginary factors multiply. Further powers cycle: i³ = −i and i⁴ = 1. A purely imaginary non-zero number has x = 0 and y ≠ 0. Zero has both parts zero; it is not a non-zero purely imaginary number. Complex numbers do not have the usual order relation of the real number line, so a comparison such as z > 0 needs a stated real quantity, such as its modulus, rather than an unstated ordering.
Equality requires both parts to match
For real x, y, a and b, x + iy = a + ib if and only if x = a and y = b. Apply this after collecting the expression into Cartesian form. If (a + 2) + (b − 1)i = 5 − 3i, real parts give a + 2 = 5 and imaginary parts give b − 1 = −3, hence a = 3 and b = −2. Substitution checks both equations at once.
If each part contains both parameters, equating parts gives simultaneous equations. For (a + b) + (a − b)i = 6 + 2i, use a + b = 6 and a − b = 2 to obtain a = 4 and b = 2. Treating i as an adjustable real variable would destroy the unique two-part representation. State that the unknown parameters are real when using this method; complex parameters would require further separation.
Pause and explain
What is the imaginary part of 7 − 2i?
Worked example
For real a and b, solve (a + 2) + (b − 1)i = 5 − 3i.
Show the worked solution
- Equate real parts: a + 2 = 5.
- Equate imaginary coefficients: b − 1 = −3.
- Hence a = 3 and b = −2; substitution returns 5 − 3i.
Answer a = 3, b = −2.
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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.
For real a and b, solve (2a − 1) + (b + 2)i = 7 − i.
Give me a hint
Write one equality for each corresponding real coefficient.
Compare my reasoning
- Real parts give 2a − 1 = 7, hence a = 4.
- Imaginary parts give b + 2 = −1, hence b = −3.
- These values give 7 − i when substituted into the original expression.
a = 4, b = −3.
Look for these in your work
- Equate both parts with their signs.
- Substitute the two parameters into the original equality.
For real a and b, solve (a + b) + (a − b)i = 6 + 2i.
Give me a hint
The two parts give a pair of simultaneous linear equations.
Compare my reasoning
- Use a + b = 6 and a − b = 2.
- Adding gives 2a = 8, so a = 4; then b = 2.
- The values reproduce real part 6 and imaginary part 2.
a = 4, b = 2.
Look for these in your work
- Retain both simultaneous equations.
- Verify both real and imaginary coefficients.
A supplied encoding of two real coordinates is z = (x + 2y) + (2x − y)i. Decode z = 7 + 4i.
Give me a hint
The two encoded parts are linear combinations of the real coordinates.
Compare my reasoning
- Equate parts: x + 2y = 7 and 2x − y = 4.
- Twice the first equation minus the second gives 5y = 10, so y = 2 and x = 3.
- The coordinates encode as (3 + 4) + (6 − 2)i = 7 + 4i.
x = 3, y = 2.
Look for these in your work
- Use the supplied encoding rather than reading its parts directly as x and y.
- Check the reconstructed complex value.
Common mistakes
- Im z is the real coefficient of i, not the entire imaginary term.
- Both parts must match; one equation does not establish equality.
- The parameters must be real for direct equating of these two parts.
What can you explain now?
What are Re(4 − 3i) and Im(4 − 3i)?
Compare with the explanation
Re z = 4; Im z = −3.
The imaginary part is the signed real coefficient of i.
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Make it stick
Define i, identify Re z and Im z, and use equality of both parts to recover real parameters.
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.