Conjugate Root Theorem
When a polynomial has real coefficients, the complex roots refuse to come alone — they always show up in conjugate pairs. This single fact lets you recover missing roots from a partial factorisation, build polynomials from prescribed complex zeros, and factor any real polynomial as a product of real linear and real quadratic pieces. This lesson proves the theorem, then turns it into a toolkit.
If $z = 3 - 2i$, write $\bar{z}$ and compute $z + \bar{z}$ and $z\bar{z}$. Before checking — both sums should be real. Why does that matter when building a polynomial with real coefficients? Sketch your reasoning below.
Two habits unlock every conjugate-root question: check the coefficients are real (otherwise the theorem doesn't apply), then pair $z$ with $\bar{z}$ and multiply them into a real quadratic $z^2 - 2\,\text{Re}(z)\,z + |z|^2$. Once you have that quadratic, polynomial division finishes the job.
The pair-multiply-divide routine: (1) pair the given complex root with its conjugate, (2) multiply $(z-\alpha)(z-\bar{\alpha})$ into a real quadratic, (3) divide the original polynomial by that quadratic to expose the remaining factor.
$(z - \alpha)(z - \bar{\alpha}) = z^2 - (\alpha + \bar{\alpha})z + \alpha\bar{\alpha} = z^2 - 2\,\text{Re}(\alpha)\,z + |\alpha|^2$
Key facts
- Conjugate Root Theorem: if $P$ has real coefficients and $P(\alpha) = 0$, then $P(\bar{\alpha}) = 0$
- $(z-\alpha)(z-\bar{\alpha}) = z^2 - 2\,\text{Re}(\alpha)z + |\alpha|^2$ is always real
- An odd-degree real polynomial has at least one real root
- Conjugation is a ring homomorphism: $\overline{z+w} = \bar{z}+\bar{w}$, $\overline{zw} = \bar{z}\bar{w}$
Concepts
- Why $P(\bar{z}) = \overline{P(z)}$ when coefficients are real
- Why this forces non-real roots to come in conjugate pairs
- Why every real polynomial factors into real linear and real quadratic pieces
Skills
- Find missing roots of a real polynomial given one complex root
- Construct a real polynomial with prescribed complex zeros
- Factor a real polynomial completely over $\mathbb{R}$
Theorem. Let $P(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0$ with every $a_k \in \mathbb{R}$. If $\alpha \in \mathbb{C}$ is a root of $P$, then $\bar{\alpha}$ is also a root.
Proof outline. Conjugation respects $+$ and $\times$, so for any $z$:
- $\overline{P(z)} = \overline{a_n z^n + \cdots + a_0} = \overline{a_n}\,\overline{z}^n + \cdots + \overline{a_0}$.
- Each $a_k$ is real, so $\overline{a_k} = a_k$. Hence $\overline{P(z)} = a_n \bar{z}^n + \cdots + a_0 = P(\bar{z})$.
- Apply this with $z = \alpha$: $P(\bar{\alpha}) = \overline{P(\alpha)} = \overline{0} = 0$. So $\bar{\alpha}$ is a root. $\blacksquare$
Worked through the hook: $P(z) = z^3 - 5z^2 + 11z - 15$. Coefficients $1, -5, 11, -15$ are all real, and $z = 2 + i$ is a root. By the theorem, $z = 2 - i$ is also a root. Multiplying gives the real quadratic $(z - (2+i))(z - (2-i)) = z^2 - 4z + 5$. Polynomial division yields $P(z) = (z^2 - 4z + 5)(z - 3)$, so the third root is $z = 3$.
Theorem: real coefficients $+$ root $\alpha$ $\Rightarrow$ root $\bar{\alpha}$ · Proof key step: $\overline{P(z)} = P(\bar{z})$ when coefficients are real · Conjugate pair gives real quadratic $z^2 - 2\,\text{Re}(\alpha)z + |\alpha|^2$ · Real polynomial = product of real linears and irreducible real quadratics
Pause — copy the conjugate root theorem statement, the proof key step $\overline{P(z)} = P(\bar{z})$, and the real quadratic factor $z^2-2\operatorname{Re}(\alpha)z+|\alpha|^2$ into your book.
Quick check: $P(z)$ has real coefficients and degree 4. Three of its roots are $1, \; 2+3i, \; -i$. What is the fourth root?
We just saw the conjugate root theorem: real-coefficient $P(z)$ has $\alpha$ as a root $\Rightarrow$ $\bar\alpha$ is also a root, proved via $\overline{P(z)} = P(\bar{z})$ for real coefficients. That raises a question: how do we construct the full real polynomial when given a mix of real and complex roots? This card answers it → pair each non-real root with its conjugate to form real quadratics $z^2 - 2\operatorname{Re}(\alpha)z + |\alpha|^2$, then multiply all factors.
To construct a real polynomial with a prescribed complex root $\alpha$, you must also include $\bar{\alpha}$. Multiply the linear factors together to absorb the imaginary parts into a single real quadratic:
- $(z - \alpha)(z - \bar{\alpha}) = z^2 - (\alpha + \bar{\alpha})z + \alpha\bar{\alpha} = z^2 - 2\,\text{Re}(\alpha)z + |\alpha|^2$.
- Both $2\,\text{Re}(\alpha)$ and $|\alpha|^2 = \alpha\bar{\alpha}$ are real, so the resulting quadratic has real coefficients.
Recipe. Given required roots (some real, some non-real), pair each non-real root with its conjugate, multiply each pair into a real quadratic, multiply by each real linear factor, and finally multiply by any leading coefficient required.
Real polynomial from roots: pair each non-real root with $\bar{\alpha}$ · Pair multiplies to $z^2 - 2\,\text{Re}(\alpha)z + |\alpha|^2$ (real quadratic) · Final $P(z)$ = (leading coeff) $\times \prod$ real quadratics $\times \prod$ real linears · If you skip the conjugate, coefficients become complex
Pause — copy the construction procedure (pair non-real roots with conjugates, form real quadratics, multiply with real linears) and the warning that skipping the conjugate produces complex coefficients into your book.
Did you get this? True or false: the monic real quadratic with roots $3 + 4i$ and $3 - 4i$ is $z^2 - 6z + 25$.
Worked examples · 3 in a row, reveal as you go
$P(z) = z^3 - 7z^2 + 17z - 15$ has real coefficients and $z = 2 - i$ is a root. Find the remaining roots and factor $P(z)$ completely over $\mathbb{R}$.
Find the monic polynomial of least degree with real coefficients having roots $z = 1$ and $z = 2 + i$.
Factor $P(z) = z^4 - 4z^3 + 14z^2 - 4z + 13$ completely over $\mathbb{R}$, given that $z = i$ is a root.
Fill the gap: If $P(z)$ has real coefficients and $\alpha = a + bi$ is a root, then $(z - \alpha)(z - \bar{\alpha}) = z^2 - \,z + $ is an irreducible real quadratic factor of $P(z)$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: every polynomial of degree 5 with real coefficients must have at least one real root.
Activities · practice with the ideas
$P(z) = z^3 - 4z^2 + 6z - 4$ has real coefficients and one root $z = 1 + i$. Find all roots and the real factorisation.
Find the monic polynomial of least degree with real coefficients having roots $z = -2$ and $z = 1 - 3i$.
Show that $P(z) = z^4 + 4$ factorises over $\mathbb{R}$ as $(z^2 + 2z + 2)(z^2 - 2z + 2)$, and hence find all four complex roots.
Explain why a degree-7 polynomial with real coefficients must have at least one real root. Could it have exactly two real roots?
Counterexample challenge: find a polynomial with complex coefficients that has a non-real root $\alpha$ but does not have $\bar{\alpha}$ as a root. Explain why this does not contradict the theorem.
Odd one out: Three of these statements are true for every polynomial $P(z)$ with real coefficients. Which one is NOT?
Earlier you analysed $P(z) = z^3 - 5z^2 + 11z - 15$ with the root $z = 2 + i$, and were asked for a second root without computation.
The second root is $z = 2 - i$, forced by the Conjugate Root Theorem because every coefficient of $P$ is real. Multiplying the pair gives the real quadratic $z^2 - 4z + 5$; dividing exposes the third (real) root $z = 3$. Recognising that "real coefficients" instantly delivers a second root for free is the most efficient move in any Module 13 question of this type.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. $P(z) = z^3 - 6z^2 + 13z - 10$ has real coefficients and $z = 2 + i$ is a root. State the second complex root, give a reason, and find the remaining root. (2 marks)
Q2. Find the monic polynomial of least degree with real coefficients having roots $z = -1$ and $z = 3 + 2i$. (3 marks)
Q3. Prove the Conjugate Root Theorem: if $P(z)$ has real coefficients and $P(\alpha) = 0$, then $P(\bar{\alpha}) = 0$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. Conjugate $1 - i$ also a root. Real quadratic $(z - (1+i))(z - (1-i)) = z^2 - 2z + 2$. Dividing $z^3 - 4z^2 + 6z - 4$ by $z^2 - 2z + 2$ gives $z - 2$. So $P(z) = (z^2 - 2z + 2)(z - 2)$, with roots $1 \pm i$ and $2$.
2. Conjugate $1 + 3i$ must be added. Pair $(z - (1-3i))(z - (1+3i)) = z^2 - 2z + 10$. Multiply by $(z + 2)$: $z^3 - 2z^2 + 10z + 2z^2 - 4z + 20 = z^3 + 6z + 20$.
3. $(z^2 + 2z + 2)(z^2 - 2z + 2) = z^4 - 2z^3 + 2z^2 + 2z^3 - 4z^2 + 4z + 2z^2 - 4z + 4 = z^4 + 4$ ✓. Roots of $z^2 + 2z + 2$ are $-1 \pm i$; roots of $z^2 - 2z + 2$ are $1 \pm i$.
4. Non-real roots come in conjugate pairs, contributing an even count. Degree 7 is odd, so the number of real roots is odd $\Rightarrow$ at least 1. Two real roots would make the non-real count $7 - 2 = 5$, which is odd, impossible. So 1, 3, 5 or 7 real roots — never exactly 2.
5. $P(z) = z - i$ has root $z = i$, and $-i$ is not a root. No contradiction: the constant coefficient $-i$ is non-real, so the theorem does not apply.
Q1 (2 marks): Second root $z = 2 - i$ by Conjugate Root Theorem (coefficients real) [1]. Real quadratic $z^2 - 4z + 5$; dividing gives $P(z) = (z^2 - 4z + 5)(z - 2)$, third root $z = 2$ [1].
Q2 (3 marks): Conjugate $3 - 2i$ also a root [1]. Pair $(z - (3+2i))(z - (3-2i)) = z^2 - 6z + 13$ [1]. Multiply by $(z + 1)$: $P(z) = z^3 - 5z^2 + 7z + 13$ [1].
Q3 (3 marks): Conjugation preserves sum and product, so $\overline{P(z)} = \overline{a_n}\bar{z}^n + \cdots + \overline{a_0}$ [1]. Each $a_k$ real means $\overline{a_k} = a_k$, hence $\overline{P(z)} = P(\bar{z})$ [1]. Set $z = \alpha$: $P(\bar{\alpha}) = \overline{P(\alpha)} = \overline{0} = 0$, so $\bar{\alpha}$ is a root [1].
Five timed questions on conjugate pairing, real quadratic factors and complete real factorisation. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick conjugate-root questions. Lighter alternative to the boss.
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