H2 Maths Conjugate Roots: Check Every Coefficient Before Pairing

A public SGExams discussion of H2 Maths revision includes a compact piece of complex-number advice: check for real coefficients before using a conjugate root. That condition deserves its own worked exercise. Without it, a useful shortcut can become an unjustified extra root.

The MOE H2 Mathematics syllabus 9758 includes conjugate roots of polynomial equations with real coefficients. The examples here use Cartesian form and polynomial factorisation. They are original exercises, rather than reconstructions of any school or national examination question.

State the condition before the conclusion

If a polynomial has real coefficients and a non-real root a + bi, where a and b are real and b is non-zero, then a − bi is also a root. Check every coefficient, including the constant term. A missing term has coefficient zero, which is real.

Real coefficients need not be positive integers. Fractions, negative numbers and real irrational numbers also qualify. Conversely, a polynomial is not automatically covered just because most of its coefficients are real. One non-real coefficient is enough to prevent you from invoking this rule without further justification.

It helps to write a short reason immediately before listing the conjugate: all coefficients are real. That sentence makes your use of the theorem inspectable. It also prompts you to return to the question when a parameter has not yet been identified as real.

Form a quadratic factor from an original root

Suppose a monic cubic with real coefficients has roots 3 + 2i and −1. The conjugate-root condition gives a third root, 3 − 2i. The factor associated with the conjugate pair is (z − 3 − 2i)(z − 3 + 2i).

Group the expression as (z − 3)² − (2i)². Since i² = −1, this becomes (z − 3)² + 4 = z² − 6z + 13. Notice the positive 4: losing the minus sign in i² would change the constant term and destroy the required roots.

The complete cubic is therefore (z + 1)(z² − 6z + 13). Expanding gives z³ − 5z² + 7z + 13. Keeping the factors until the last step makes the origin of each root visible, rather than hiding the structure in a long complex expansion.

Verify the result in two ways

Substitute z = 3 + 2i into the quadratic factor. Here (3 + 2i)² = 5 + 12i, so z² − 6z + 13 becomes (5 + 12i) − (18 + 12i) + 13 = 0. The conjugate gives zero as well. Substituting −1 into the cubic gives −1 − 5 − 7 + 13 = 0.

You can also inspect the conjugate pair's sum and product. Their sum is 6 and their product is 13. A monic quadratic with those two roots is therefore z² − 6z + 13, matching the factor above. This is a useful independent sign check.

Recognise an example where pairing fails

Consider the original polynomial (z − i)(z − 2) = z² − (2 + i)z + 2i. Its roots are i and 2. Although i is a root, −i is not. Direct substitution gives −2 + 4i, which is non-zero.

The polynomial has non-real coefficients, so the real-coefficient condition was never satisfied. This example does not say that a polynomial with complex coefficients can never have a conjugate pair. It says the pair cannot be inferred from this theorem alone. You would need to calculate or use other information.

Turn the condition into a revision habit

Before solving a polynomial question, circle any statement about the coefficients and underline any statement about a parameter being real. Then list the roots actually supplied separately from roots you have deduced. This prevents an assumption from looking like given information halfway through a solution.

Choose a relevant question from the JC prelim paper collection. Ask whether a conjugate supplies a new root, whether the resulting factor has real coefficients and whether substitution confirms it. When practising with older papers, compare their content with your own syllabus before spending time on unfamiliar methods. One thoroughly checked factorisation is more useful than several answers whose coefficient condition was never examined.