IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 HL only ~10 min read

Complex Roots of Polynomials

Now that you know how to handle complex numbers, here’s a satisfying payoff: every polynomial equation has a full set of solutions, as long as you allow complex numbers. A quadratic with no real roots? It’s got two complex ones. A cubic that only crosses the x-axis once? The two missing roots are complex, and they come as a pair. The pattern that makes all this work is one of the prettiest results in HL maths: when a polynomial has real coefficients, complex roots always show up in conjugate pairs. Once you trust this rule, solving polynomials becomes a clean, step-by-step process — even when imaginary numbers are hiding in the middle of the equation.

📘 What you need to know

A quadratic az² + bz + c = 0 (with real coefficients) has complex roots when the discriminant b² − 4ac is negative.
Solve a quadratic with complex roots using the quadratic formula or by completing the square — the same techniques you already know, just with √(negative).
For a polynomial with real coefficients, complex roots always come in conjugate pairs: if p + qi is a root, so is p − qi.
This is only true when all coefficients are real. If any coefficient is complex, the conjugate-pair rule breaks.
A polynomial of degree n has exactly n roots (counting repeats). Some may be real, others complex — but the total is always n.
Given one complex root of a cubic, find the conjugate (= a second root), build the quadratic factor (z − z₁)(z − z₂), then divide the cubic by it to get the linear factor.
Useful expansion: (z − (p+qi))(z − (p−qi)) = z² − 2pz + (p² + q²) — this gives a real quadratic factor every time.

When does a quadratic have complex roots?

You already know that a quadratic az² + bz + c = 0 has its discriminant Δ = b² − 4ac. The three cases:

Discriminant decides the root types Δ > 0 → two distinct real roots
Δ = 0 → one real repeated root
Δ < 0 → two complex conjugate roots

The negative-discriminant case is the new territory. Before complex numbers, you’d write “no real solutions” and stop. Now you finish the job using i.

A handy check: substitute your complex roots back into the original equation. If everything cancels to 0 + 0i, you’ve solved it correctly. If you get something else, there’s an arithmetic slip somewhere.

The conjugate-pair rule

Here’s the rule that makes complex roots tractable. Whenever a polynomial has real coefficients, any complex root brings its conjugate along for the ride.

Complex Conjugate Root Theorem if z₁ = p + qi is a root of a polynomial with real coefficients,
then z₂ = p − qi is also a root.

Complex roots come in mirror-image pairs

🤔 Why must complex roots come in pairs?

Think of a polynomial P(z) with real coefficients. If you plug in z = p + qi and the result is 0, then taking the complex conjugate of both sides gives P(p − qi) = 0 too — because the conjugate of any real coefficient is itself, so the whole polynomial structure is preserved. The conjugate has to be a root if the original is.

Important caveat: the conjugate-pair rule only works when all coefficients of the polynomial are real. If a coefficient itself is complex (e.g. (3 + i)z² + …), the rule fails — complex roots may appear without their conjugates.

How many roots does a polynomial have?

The Fundamental Theorem of Algebra tells you exactly: a polynomial of degree n has exactly n roots in the complex numbers, counting repeated roots separately. Some are real, others complex — but the total always equals the degree.

Counting roots by degree degree 2 (quadratic) → 2 roots
degree 3 (cubic) → 3 roots
degree 4 (quartic) → 4 roots
degree n → n roots

For a polynomial with real coefficients, since complex roots come in pairs, the number of complex roots must be even. Combine this with the total count, and you can predict the structure straight from the degree:

Cubic (degree 3, real coefficients): either 3 real roots or 1 real root + 1 complex conjugate pair. Never two complex roots without a third — the pair always comes together.

Factoring a quadratic with complex roots

Once you’ve solved a quadratic and found its complex roots z₁ and z₂, you can write the quadratic in factored form:

Factored form az² + bz + c = a(z − z₁)(z − z₂)

If the leading coefficient a = 1, this becomes simply (z − z₁)(z − z₂). Multiplying out the conjugate-pair factors gives a beautiful identity:

Conjugate-pair expansion (z − (p + qi))(z − (p − qi)) = z² − 2pz + (p² + q²)

Notice that the answer is a quadratic with real coefficients. That’s no coincidence — the imaginary parts cancel by the difference-of-squares pattern. This identity is gold: any time you have two conjugate roots, you can build their real quadratic factor in one line.

Building a quadratic when you’re given one root

A common exam type: “given that p + qi is a root of z² + Bz + C = 0, find the values of B and C“. The recipe:

🧭 Recipe — find a quadratic from one complex root

Use the conjugate-pair rule — if p + qi is a root, so is p − qi.
Multiply the factors (z − (p+qi))(z − (p−qi)) using the identity from above.
Read off B = −2p and C = p² + q².

Solving a cubic given one root

Cubics with real coefficients can be cracked in two situations: when you’re given a real root, or when you’re given a complex one. Either way, the approach is to find a factor and divide it out.

🧭 Recipe — solve a cubic given one complex root

Use the conjugate to find the second root — if p + qi is a root, then p − qi is too.
Build the quadratic factor using (z − (p+qi))(z − (p−qi)) = z² − 2pz + (p² + q²).
Divide the cubic by the quadratic factor (using polynomial division or comparing coefficients).
Solve the linear quotient to find the third (real) root.

Coefficient-comparison shortcut: instead of doing polynomial division, you can multiply (az + b)(z² + Bz + C) and equate with the original cubic to find the linear factor’s coefficients quickly. Often faster than long division.

Higher-degree polynomials

The same logic extends to quartics (degree 4), quintics, and beyond. If you’re given two complex roots of a quartic, the conjugate-pair rule gives you four roots total — so you can immediately build two quadratic factors and multiply them together.

For example, if 1 + i and 4 + 5i are both roots of a quartic with real coefficients, then 1 − i and 4 − 5i are also roots. The quartic factors as:

(z − (1+i))(z − (1−i))(z − (4+5i))(z − (4−5i))
= (z² − 2z + 2)(z² − 8z + 41)

Worked examples

WE 1

Solve a quadratic with complex roots

Solve z² − 4z + 13 = 0, giving your answers in Cartesian form.

Step 1: Identify a, b, c a = 1, b = −4, c = 13 Step 2: Discriminant check b² − 4ac = 16 − 52 = −36 (negative → complex roots) Step 3: Apply the quadratic formula z = (4 ± √(−36)) / 2 √(−36) = 6i z = (4 ± 6i) / 2 = 2 ± 3i z₁ = 2 + 3i, z₂ = 2 − 3i notice the roots are conjugates of each other — guaranteed because the coefficients are real

WE 2

Factor a quadratic into complex linear factors

Factor z² − 6z + 25 into the form (z − z₁)(z − z₂) where z₁, z₂ are complex.

Step 1: Solve to find the roots z = (6 ± √(36 − 100)) / 2 = (6 ± √(−64)) / 2 = (6 ± 8i) / 2 = 3 ± 4i Step 2: Write the factored form using these roots z₁ = 3 + 4i, z₂ = 3 − 4i z² − 6z + 25 = (z − 3 − 4i)(z − 3 + 4i) check: expanding gives z² − (3+4i+3−4i)z + (3+4i)(3−4i) = z² − 6z + (9 + 16) = z² − 6z + 25 ✓

WE 3

Find the quadratic from one complex root

Given that 1 + 4i is a root of z² + Bz + C = 0, where B, C ∈ ℝ, find the values of B and C.

Step 1: Conjugate-pair rule — second root is 1 − 4i factors: (z − (1+4i))(z − (1−4i)) Step 2: Apply the conjugate-pair identity (with p=1, q=4) = z² − 2(1)z + (1² + 4²) = z² − 2z + 17 Step 3: Compare with z² + Bz + C B = −2, C = 17 B = −2, C = 17 the identity z² − 2pz + (p²+q²) is your friend — it skips the manual expansion

WE 4

Solve a cubic given one complex root

Given that 3 + 2i is a root of z³ − 5z² + 19z − 39 = 0, find the other two roots.

Step 1: Conjugate is also a root second root: z₂ = 3 − 2i Step 2: Build quadratic factor (p=3, q=2) z² − 2(3)z + (3² + 2²) = z² − 6z + 13 Step 3: Divide cubic by z² − 6z + 13 to get linear factor (az + b) (az + b)(z² − 6z + 13) = z³ − 5z² + 19z − 39 comparing z³ coefficient: a = 1 comparing constant: 13b = −39 → b = −3 linear factor: (z − 3) Step 4: Solve linear factor z − 3 = 0 → z = 3 roots: 3 + 2i, 3 − 2i, 3 comparing leading and constant coefficients is faster than full polynomial long division

WE 5

Quartic with two complex roots given

Given that 2 + i and 1 − 3i are both roots of a quartic P(z) with real coefficients and leading coefficient 1, write down all four roots and find P(z).

Step 1: Conjugate-pair rule gives the other two roots 2 + i is a root → 2 − i is also a root 1 − 3i is a root → 1 + 3i is also a root Step 2: Build two quadratic factors using the identity first pair (p=2, q=1): z² − 4z + (4 + 1) = z² − 4z + 5 second pair (p=1, q=3): z² − 2z + (1 + 9) = z² − 2z + 10 Step 3: Multiply the two quadratics (z² − 4z + 5)(z² − 2z + 10) = z⁴ − 2z³ + 10z² − 4z³ + 8z² − 40z + 5z² − 10z + 50 = z⁴ − 6z³ + 23z² − 50z + 50 P(z) = z⁴ − 6z³ + 23z² − 50z + 50 all four roots: 2 ± i and 1 ± 3i — the conjugate rule gives you the missing two for free

WE 6

Find a missing coefficient

The cubic z³ + 4z² + kz + 50 = 0 has 1 + 3i as a root, where k ∈ ℝ. Find the value of k.

Step 1: Conjugate is also a root → 1 − 3i quadratic factor: z² − 2z + (1 + 9) = z² − 2z + 10 Step 2: Express cubic as (z² − 2z + 10)(z + c) for some real c expand: (z² − 2z + 10)(z + c) = z³ + cz² − 2z² − 2cz + 10z + 10c = z³ + (c − 2)z² + (10 − 2c)z + 10c Step 3: Compare with z³ + 4z² + kz + 50 z² coefficient: c − 2 = 4 → c = 6 constant: 10c = 50 ✓ (consistency check) z coefficient: 10 − 2c = 10 − 12 = −2 k = −2 use one comparison to find c, then plug into the z-coefficient to find k — clean and quick

💡 Top tips

Memorise the conjugate-pair identity: (z − (p+qi))(z − (p−qi)) = z² − 2pz + (p²+q²). Cuts the working in half.
Always verify the coefficients are real before applying the conjugate-pair rule. If any coefficient is complex, the rule doesn’t apply.
Use coefficient-comparison instead of polynomial long division when the cubic has a known leading coefficient — usually faster and less error-prone.
For “find the other roots” cubics, count first. A cubic has 3 roots total. If one is complex, two are complex (conjugate pair) and one is real.
Sketching helps for higher polynomials. Plot the known complex roots on an Argand diagram — the symmetry across the real axis reminds you to add the conjugates.
Always check by substitution. If a complex root really is a root, plugging it back into the polynomial should give 0 + 0i.
Quartics with two given complex roots are usually quartic = (real quadratic 1) × (real quadratic 2). Build each quadratic from a conjugate pair and multiply.

⚠ Common mistakes

Forgetting to add the conjugate as another root. If you find one complex root in a real-coefficient polynomial, the conjugate is automatic — don’t leave it out.
Applying the conjugate-pair rule when a coefficient is complex. The rule only holds when ALL coefficients are real. Re-read the question.
Sign errors when expanding (z − (p + qi))(z − (p − qi)). The −2pz is negative; the constant p² + q² is positive (because q² · i² = −q², which then subtracts to add).
Treating p² + q² as p² − q². The i² flips the sign of −q²·i² to +q², so the quadratic constant is always +(p² + q²).
Forgetting that a cubic has 3 roots total. If the conjugate pair only accounts for 2, a third (real) root must exist — find it via the linear factor.
Skipping the discriminant check. If Δ ≥ 0, the roots are real — no complex roots are involved at all.
Mis-identifying the linear factor. When dividing a cubic by a quadratic, the quotient is a degree-1 polynomial of the form az + b. If the cubic is monic, a = 1.

Complex roots of polynomials are where complex numbers earn their place in algebra. The conjugate-pair rule is the single most useful fact in this section — it turns “we’re given one root” into “we now have two roots”, which is enough to crack most exam questions in one or two steps. The next note brings out one of the most powerful results in HL maths: De Moivre’s theorem, which lets you take any power of a complex number in a single line, no matter how large the exponent.

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