Did you know? One of the most useful shortcuts in JC1 H2 Mathematics comes from a simple pair: (a + bi)(a − bi) = a² + b². The imaginary parts disappear because they occur with opposite signs, while i² = −1 turns the final negative square into a positive contribution. That is why a complex conjugate can turn a complex denominator into an ordinary real number.
For students looking for a JC1 H2 Mathematics Tutor Bukit Timah, this is an important foundation in the Singapore–Cambridge A-Level H2 Mathematics syllabus 9758. The 2027 syllabus includes complex numbers in Cartesian form, their four operations, conjugates, modulus, argument and Argand diagrams. It excludes complex-number polar and exponential forms. Dividing a complex number and expressing the answer clearly as x + yi is therefore a syllabus-relevant skill.
This guide gives a quick method first, followed by the reason it works, several exact examples, an Argand interpretation, zero-denominator checks, common mistakes, a practice set and a way for tutors and parents to check whether the idea is understood. It focuses on complex division, not a duplicate explanation of conjugate roots of real-coefficient polynomials.
Quick answer: multiply by the conjugate of the denominator
To divide one complex number by another, we want the answer in Cartesian form, such as 2 − 3i, with a real number in the denominator if we use a fraction.
If the denominator is c + di, its conjugate is c − di. Multiply the numerator and denominator by this conjugate, provided c and d are not both zero:
(a + bi)/(c + di) = [(a + bi)(c − di)]/(c² + d²).
The new denominator is real because (c + di)(c − di) = c² + d².
Expand the numerator, use i² = −1 and collect the real and imaginary parts:
(a + bi)/(c + di) = [(ac + bd) + (bc − ad)i]/(c² + d²).
You do not need to memorise that final expression as a separate trick if you understand the conjugate method.
- Identify the denominator’s real and imaginary parts.
- Change only the imaginary sign to form its conjugate.
- Multiply the numerator and denominator by the same non-zero complex number.
- Expand carefully and replace i² by −1.
- Collect the real terms and imaginary terms.
- Divide both collected coefficients by the real denominator.
- Check the result by multiplying it by the original denominator.
This method is reliable because every step is an equality. We are changing the form of the fraction, not changing its value.
What is a complex conjugate?
A complex number has Cartesian form z = a + bi, where a and b are real numbers and i² = −1. The number a is its real part, and b is the coefficient of its imaginary part.
The complex conjugate of z is written z̄ and equals a − bi. Only the sign of the imaginary part changes. The real part stays a.
For example, the conjugate of 3 + 4i is 3 − 4i. The conjugate of −2 − 5i is −2 + 5i. The conjugate of 7 is 7, because its imaginary coefficient is zero.
A common mistake is to change both signs. The conjugate of 3 + 4i is not −3 − 4i. That would be the negative of the original number, a different operation.
On an Argand diagram, conjugation reflects a point across the real axis. The point representing 3 + 4i is (3, 4), and its conjugate is (3, −4). Their distances from the origin are equal.
Why the product of conjugates is real
Start from z = a + bi and its conjugate z̄ = a − bi.
Multiply:
(a + bi)(a − bi) = a² − abi + abi − b²i².
The middle terms cancel. Since i² = −1, the last term becomes −b²(−1) = +b². Thus
z z̄ = a² + b².
Because a and b are real, the result is a non-negative real number. It is zero only when both a and b are zero, meaning z = 0.
For a non-zero complex number, a² + b² is strictly positive. So it is safe to use as a denominator after multiplying by a conjugate.
This quantity also equals |z|², the square of the modulus of z. The modulus |z| is the distance from the Argand point (a, b) to the origin: √(a² + b²).
The geometry and algebra tell the same story. Conjugation reflects the point but does not change its distance from the origin; multiplying z by z̄ removes the angular part and leaves the squared length.
Worked example 1: dividing by 1 − i
Calculate (3 + 2i)/(1 − i) and express the answer as a + bi.
The denominator is 1 − i. Its conjugate is 1 + i. Multiply top and bottom by 1 + i:
(3 + 2i)/(1 − i) = [(3 + 2i)(1 + i)]/[(1 − i)(1 + i)].
Expand the numerator:
(3 + 2i)(1 + i) = 3 + 3i + 2i + 2i² = 3 + 5i − 2 = 1 + 5i.
The denominator is 1 − i² = 1 − (−1) = 2.
Therefore the final answer is
1/2 + (5/2)i.
Check by multiplying the result by 1 − i:
(1/2 + (5/2)i)(1 − i) = 1/2 − (1/2)i + (5/2)i − (5/2)i² = 1/2 + 2i + 5/2 = 3 + 2i.
We recover the original numerator, so the division is correct.
Why the check matters
A student can make a small sign error in a long expansion and still obtain an answer in perfectly acceptable a + bi form. A neat final line is not proof that the numerator has been expanded correctly.
The multiplication check is fast. Take your proposed quotient, multiply by the original denominator and see whether it gives the original numerator.
That check is especially valuable when i² appears and the student has been tempted to use i² = +1 by accident.
Worked example 2: dividing by 2 + 3i
Now calculate (1 + i)/(2 + 3i).
Use the conjugate of the denominator, 2 − 3i:
(1 + i)/(2 + 3i) = [(1 + i)(2 − 3i)]/[(2 + 3i)(2 − 3i)].
Expand the numerator:
(1 + i)(2 − 3i) = 2 − 3i + 2i − 3i² = 2 − i + 3 = 5 − i.
The denominator is 2² + 3² = 13.
So the answer is
5/13 − (1/13)i.
Notice that the denominator is not 2² − 3² = −5. Because i² = −1, the conjugate product is a sum of squares, not a difference of ordinary positive squares.
To check, multiply (5 − i)/13 by 2 + 3i:
(5 − i)(2 + 3i) = 10 + 15i − 2i − 3i² = 13 + 13i.
Divide by 13 and recover 1 + i.
Worked example 3: the answer can become a neat integer pair
Evaluate (4 − 3i)/(2 + i).
The conjugate of 2 + i is 2 − i. Multiply numerator and denominator by 2 − i.
The numerator is
(4 − 3i)(2 − i) = 8 − 4i − 6i + 3i² = 8 − 10i − 3 = 5 − 10i.
The denominator is (2 + i)(2 − i) = 4 + 1 = 5.
Thus
(4 − 3i)/(2 + i) = 1 − 2i.
Verify:
(1 − 2i)(2 + i) = 2 + i − 4i − 2i² = 2 − 3i + 2 = 4 − 3i.
The quotient is correct.
This example illustrates why multiplying by a conjugate is useful even if we expect a simple result. Without the method, a student might try to “divide the real parts and imaginary parts separately,” which is not a valid operation.
A false shortcut: separate the real and imaginary parts of the fraction
Consider the expression (4 − 3i)/(2 + i). A wrong move is to write 4/2 − (3/1)i = 2 − 3i.
That has no general algebraic justification. The denominator 2 + i is one complex number, not two unrelated divisors. Dividing by a sum does not mean dividing each term of the numerator by the matching term of the denominator.
You can disprove the shortcut instantly by multiplying the proposed answer 2 − 3i by 2 + i. It does not return 4 − 3i.
Complex division works through ordinary field operations. We can multiply numerator and denominator by the same non-zero number; we cannot split one fraction across its denominator’s two terms.
The conjugate method is not a special exemption. It is exactly the legal fraction operation that makes further simplification possible.
How this connects to familiar surds
Students may have seen rationalisation at O-Level Additional Mathematics: for example, multiply 1/(2 + √3) by (2 − √3)/(2 − √3) to make the denominator real and simplify it.
The complex conjugate method has a similar pattern. Multiplying 1/(2 + 3i) by (2 − 3i)/(2 − 3i) produces a real denominator.
But the numerical identity is different. With real surds, (2 + √3)(2 − √3) = 4 − 3 = 1. With the imaginary term, (2 + 3i)(2 − 3i) = 4 − (3i)² = 4 + 9 = 13.
Both methods use conjugate pairs, but i² = −1 changes the final sign. A useful tutor will highlight the similarity and the difference rather than expecting a student to copy the surd method without thought.
What happens when the denominator is zero?
Every fraction requires a non-zero denominator, including a complex fraction.
If c + di = 0, then its real and imaginary parts must both be zero: c = 0 and d = 0. There is no legal division by that number.
The conjugate product c² + d² is also zero in exactly that case. Multiplying by the conjugate cannot fix division by zero. You are not allowed to divide by zero and then rationalise it away.
When the denominator is non-zero, c² + d² is positive, so the rationalisation method is well-defined.
The correct habit is the same as in real-number algebra: check whether the original denominator can be zero before simplifying.
Complex conjugation on the Argand diagram
A complex number a + bi corresponds to the point (a, b) on the Argand plane. The horizontal axis shows the real part, and the vertical axis shows the imaginary part.
Changing a + bi to a − bi reflects the point across the horizontal real axis. The distance from the origin remains unchanged because the two points have the same squared coordinates a² + b².
For z = 3 + 4i, both z and z̄ have modulus five. Their product is 25, equal to the modulus squared.
This visual interpretation is useful because students sometimes confuse conjugation with negation. Negating 3 + 4i gives −3 − 4i, which reflects through the origin, whereas conjugation gives 3 − 4i, a reflection across the real axis.
The 2027 syllabus includes graphical effects of conjugation, negation and other basic operations in Cartesian form. Polar and exponential complex-number forms are excluded, so a good H2 guide need not rely on advanced notation to explain the core division method.
Why conjugate roots are a related but different question
For a polynomial with real coefficients, if a non-real complex number a + bi is a root, its conjugate a − bi is also a root. That property is related to conjugation but is not the same task as dividing one complex number by another.
In division, we deliberately multiply by a conjugate to create a real denominator. In a polynomial-root question, the coefficients and the conjugation property explain why roots come in pairs.
It is useful to recognise both applications, but students should not mix them together. Being asked to simplify (3 + 2i)/(1 − i) does not require solving a polynomial equation.
For wider context, consult the Bukit Timah H2 complex numbers and Argand diagrams guide.
A five-step tutor diagnostic
Begin with the basic question: “What is the conjugate of 4 − 7i?” The answer should be 4 + 7i. If a learner changes both signs, fix the definition before moving to division.
Next, ask them to expand (4 − 7i)(4 + 7i). The answer is 16 + 49 = 65, not 16 − 49. If the learner gets this wrong, the issue is likely i² = −1 or signs in multiplication.
Then give (1 + i)/(1 − i). Ask which expression to multiply by and why. The learner should say “1 + i, because it makes the denominator two.”
After that, use a less tidy pair such as (1 + i)/(2 + 3i) to test reliable expansion and collection. Finally, require a multiplication check of the result.
These stages are deliberately narrow. A student who can reproduce the process on a new denominator has learned a transferrable method.
Ten short practice questions with answers
Question 1. What is the conjugate of 3 + 4i? Answer: 3 − 4i.
Question 2. For z = 3 + 4i, calculate z z̄. Answer: 3² + 4² = 25.
Question 3. Simplify (1 + i)/(1 − i). Answer: Multiply by (1 + i)/(1 + i), giving (1 + i)²/2 = (2i)/2 = i.
Question 4. Simplify (5)/(2 − i). Answer: Multiply by (2 + i)/(2 + i). The result is 5(2 + i)/5 = 2 + i.
Question 5. Simplify (3 − i)/(1 + i). Answer: Multiply by (1 − i)/(1 − i). The numerator becomes 2 − 4i, and the denominator is two. Answer: 1 − 2i.
Question 6. Simplify (2 − 2i)/(1 − i). Answer: Factor the numerator as 2(1 − i) and cancel the non-zero common factor. The answer is 2.
Question 7. What is the conjugate of −4 − 3i? Answer: −4 + 3i.
Question 8. Evaluate (1 − i)(1 + i). Answer: 1 − i² = 2.
Question 9. Simplify (2 + 3i)/(2 − 3i). Answer: Multiply by 2 + 3i. The numerator becomes (2 + 3i)² = −5 + 12i and the denominator is 13. Answer: −5/13 + (12/13)i.
Question 10. Can multiplying by a conjugate make division by 0 + 0i valid? Answer: No. The original denominator is zero, and division is undefined.
These questions cover definition, multiplication, division, cancellation and the zero case. A good independent learner should also explain why every final rationalised denominator is real.
How to check the answer without a calculator
If a calculated quotient is q and the original expression is A/B, multiply q by B and see whether you recover A. This is an exact algebraic check that does not depend on memorising the conjugate formula twice.
For example we obtained (4 − 3i)/(2 + i) = 1 − 2i. Multiplying (1 − 2i)(2 + i) gave 4 − 3i. That confirms both the real and imaginary coefficients.
If the check fails, return to the numerator expansion and inspect every i² term. A sign error there is common and usually easy to correct.
The other useful check is the denominator: after multiplying by a true conjugate of a non-zero complex number, the denominator must be a² + b², a positive real number. If an imaginary term remains, the conjugate was wrong or the product was expanded incorrectly.
Frequently asked questions
Is the complex conjugate the same as the negative?
No. The conjugate of a + bi is a − bi. Its negative is −a − bi. These are generally different.
Why does the denominator become real?
Because (a + bi)(a − bi) = a² + b². The imaginary cross terms cancel and i² = −1.
Does a complex denominator mean the answer is not valid?
No. It is a valid complex fraction when the denominator is non-zero. Rationalising expresses the same value in clearer Cartesian form, often required by the question.
Must the final answer be written as a + bi?
Follow the question’s instructions. Cartesian form is a common and syllabus-relevant representation for 2027 H2 Mathematics complex-number operations.
Are polar and exponential forms required?
No. The 2027 H2 Mathematics syllabus 9758 explicitly excludes complex numbers expressed in polar or exponential form. This guide stays with Cartesian operations.
What does multiplying the quotient by the denominator prove?
If it returns the original numerator, it verifies the division identity exactly.
Related Mathematics reading and consultation
Explore the existing H2 complex numbers and Argand diagrams overview or the national H2 complex-number lesson for wider coverage. Students who need an independent problem-solving habit can also look at implicit differentiation when x and y mix and why cross-product order matters.
For current JC1 learning arrangements, use the Bukit Timah JC1 tuition route. The eduKateSG Secondary 1 small-group Mathematics teaching reference is useful for its diagnose–explain–practise–check approach, but it does not establish a JC1 timetable or class availability.
The idea to carry into the next problem is simple: a complex conjugate changes one sign, and that one sign change can turn a complicated-looking division into ordinary real-denominator algebra.