Complex Numbers · step 8 of 13

Solving polynomial equations over ℂ

Over ℂ every polynomial of degree n has exactly n roots counted with multiplicity, so a quadratic with a negative discriminant has two complex roots rather than none. When the coefficients are real, non-real roots arrive in conjugate pairs — which lets you find one root and get its partner free, then divide out the resulting real quadratic factor.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Use the quadratic formula regardless of the discriminantA negative discriminant simply produces ±i√|Δ| in the numerator.
Exploit conjugate pairs for real coefficientsKnowing one non-real root gives its conjugate immediately.
Build the real quadratic factorFrom roots a ± bi, the factor is z² − 2az + (a² + b²).
Divide out and solve what remainsPolynomial division reduces the degree; repeat until only linear or quadratic factors remain.

Worked example

Solve z² − 4z + 13 = 0, then factor z³ − 3z² + 9z + 13 given that it has a root in common with it.

Quadratic formula: z = (4 ± √(16 − 52))/2 = (4 ± √(−36))/2.
√(−36) = 6i, so z = (4 ± 6i)/2 = 2 ± 3i.
For the cubic, z² − 4z + 13 is therefore a factor.
Divide: z³ − 3z² + 9z + 13 = (z² − 4z + 13)(z + 1).

Answer. The quadratic gives 2 ± 3i; the cubic has roots 2 + 3i, 2 − 3i and −1.

Where marks get dropped

These are the specific errors that cost credit on solving polynomial equations over ℂ questions — QED's rubric penalises each of them separately.

Reporting "no solutions" for a negative discriminant. Over ℂ there are always two.
Assuming conjugate pairing when the coefficients are not all real. z² − iz = 0 has roots 0 and i, which are not conjugates.
Forgetting to count multiplicity. (z − 1)² = 0 has the single root 1 with multiplicity 2, which is still "two roots" in the degree count.

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Solving polynomial equations over ℂ — frequently asked questions

Can a real cubic have three non-real roots?

No. Non-real roots pair up, so an odd-degree real polynomial must have at least one real root — which is also why every real cubic crosses the x-axis.

How do I build a polynomial from given roots?

Multiply the linear factors. For roots 2 ± 3i, (z − 2 − 3i)(z − 2 + 3i) = (z−2)² + 9 = z² − 4z + 13.

What about repeated roots?

They count with multiplicity towards the total of n, and a repeated root is also a root of the derivative — which is how to detect them.

The rest of Complex Numbers

Arithmetic, the Argand plane, polar form and De Moivre. Each subtopic below has its own method, worked example and mark-losing traps.

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