Find both square roots with exact Cartesian working
Set a proposed root equal to a + bi, equate parts of its square and retain the two opposite roots of a non-zero complex number.
Section 3.9 · Complex numbers
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 (a + bi)² and equate real and imaginary parts.
- Solve real quadratic relations and use modulus.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the imaginary coefficient of (a + bi)²?
What you will learn
- Find both square roots of a non-zero complex number by Cartesian algebra.
- Choose linked signs correctly and verify each root by squaring.
The two equations link the signs
To solve w² = x + iy, let w = a + bi with real a and b. Expansion gives a² − b² = x and 2ab = y. Also a² + b² = √(x² + y²), since |w|² = |w²|. Adding and subtracting these real equations gives a² = [√(x² + y²) + x]/2 and b² = [√(x² + y²) − x]/2. The signs of a and b must still satisfy 2ab = y; they cannot be chosen independently.
For w² = 8 − 6i, the modulus of the target is 10, so a² = 9 and b² = 1. The condition 2ab = −6 selects (a,b) = (3,−1) or (−3,1). Thus w = 3 − i or −3 + i. Both square to 8 − 6i; 3 + i squares to its conjugate and is not a root of the stated equation.
Check both roots and special real cases
A non-zero complex number has two distinct square roots, w and −w. On an Argand diagram their points are opposite across the origin and have the same modulus. The target’s modulus is the square of either root’s modulus, so roots of 8 − 6i have modulus √10, not 10. This method shows full Cartesian working without requiring a general nth-root formula.
If y = 0, use both the square and product conditions. A positive real target has two opposite real roots; a negative real target has two opposite purely imaginary roots. For target zero, a² + b² = 0 forces a = b = 0, so there is only one distinct root, zero. Avoid dividing by a before checking a = 0, because it would lose the purely imaginary case.
Pause and explain
If a² = 9, b² = 1 and 2ab = −6, which sign pairs work?
Worked example
Find both square roots of 8 − 6i in exact Cartesian form with full working.
Show the worked solution
- Let w = a + bi; equate parts to get a² − b² = 8 and 2ab = −6.
- The target modulus is 10, so a² + b² = 10; hence a² = 9 and b² = 1.
- The product sign selects w = 3 − i and −3 + i; square both to verify.
Answer w = ±(3 − i), namely 3 − i and −3 + i.
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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.
Find both square roots of −7 + 24i by Cartesian working.
Give me a hint
Combine the real-part equation with the target modulus 25.
Compare my reasoning
- Use a² − b² = −7, 2ab = 24 and a² + b² = 25.
- Adding and subtracting give a² = 9 and b² = 16.
- Since ab = 12 is positive, the roots are 3 + 4i and −3 − 4i; both square to −7 + 24i.
±(3 + 4i).
Look for these in your work
- Show the two squared-component equations and modulus relation.
- Choose linked signs and verify both roots.
Find both square roots of −8 − 6i by exact Cartesian working.
Give me a hint
The target modulus is 10; its imaginary part fixes opposite signs.
Compare my reasoning
- Use a² − b² = −8, 2ab = −6 and a² + b² = 10.
- Hence a² = 1 and b² = 9.
- The sign condition gives roots 1 − 3i and −1 + 3i; squaring yields −8 − 6i.
±(1 − 3i).
Look for these in your work
- Use the correct real-part sign.
- Retain both roots with the required imaginary coefficient.
Solve z² = −12i in exact Cartesian form, showing the sign condition and verifying both roots.
Give me a hint
A zero real part means the two squared Cartesian components are equal.
Compare my reasoning
- Set z = a + bi: a² − b² = 0 and 2ab = −12.
- The target modulus is 12, so a² + b² = 12 and a² = b² = 6.
- Opposite signs give z = ±√6(1 − i); their squares are 6(−2i) = −12i.
z = ±√6(1 − i).
Look for these in your work
- Solve the zero-real-part case without losing the product sign.
- Verify the two opposite roots in the stated equation.
Common mistakes
- Find both distinct roots for a non-zero target.
- Link signs through 2ab = y rather than taking four independent sign combinations.
- The target modulus equals the square of the root modulus; zero has one distinct root.
What can you explain now?
How many distinct square roots does zero have?
Compare with the explanation
One: zero.
The two opposite roots coincide only for a zero target.
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Make it stick
Find both Cartesian square roots, show their linked signs, verify by squaring and distinguish the zero exception.
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.