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Module 13 · L07 of 12 ~45 min ⚡ +90 XP available

Solving Complex Polynomial Equations

When a polynomial equation has complex roots, a single clue, one root, the sum, or the product, is often enough to unlock all of them. This lesson weaves together the factor theorem, the conjugate root theorem and Vieta's formulas into one strategy for cracking polynomial equations of degree 3 and 4. Master the combination and you can solve any HSC Extension 2 polynomial in minutes.

Today's hook, The cubic $P(z) = z^3 - 5z^2 + 11z - 15$ has real coefficients and one root is $z = 1 + 2i$. Before reading on, predict (a) one other root you can name immediately, and (b) what the sum of all three roots must equal. Compare your answers after card 05.

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Recall, your gut answer first

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If a quadratic with real coefficients has one root $z = 3 - 2i$, what is the other root and what is the quadratic (with leading coefficient 1)? Before checking use the conjugate pair to write the factorisation, then expand. Sketch your reasoning below.

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Three tools, one strategy

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Solving a complex polynomial equation rests on three results you already know: the factor theorem (if $z = \alpha$ is a root then $(z - \alpha)$ is a factor), the conjugate root theorem (real-coefficient polynomials have complex roots in conjugate pairs), and Vieta's formulas (sum/product of roots come straight from the coefficients). The trick at HSC is knowing which tool to grab first.

Strategy: (1) conjugate any given complex root if coefficients are real, (2) factor out the resulting real quadratic, (3) use Vieta to cross-check sum and product against the original coefficients.

For $az^n + \ldots + k = 0$: $\sum \alpha_i = -b/a$ · $\prod \alpha_i = (-1)^n k/a$

Migrated Conjugate-factor-Vieta flow to sequence layout

$P(\alpha) = 0 \;\Rightarrow\; (z - \alpha) \mid P(z)$

Real coefficients ⇒ conjugate pairs

If $P(z)$ has all real coefficients and $z = \alpha$ is a root, then $z = \bar\alpha$ is also a root. Always check coefficients are real before invoking this.

Conjugate pair gives real quadratic

$(z - \alpha)(z - \bar\alpha) = z^2 - 2\,\text{Re}(\alpha)z + |\alpha|^2$, a quadratic with real coefficients. Divide this out of $P(z)$ to reduce the degree by 2.

Vieta for complex coefficients

Even when coefficients are not real, Vieta still applies. Given partial information (sum, product, or one root) you can recover the rest by setting up equations in the unknown roots.

What you'll master

Know

Key facts

Factor theorem: $P(\alpha) = 0 \Leftrightarrow (z - \alpha) \mid P(z)$
Conjugate root theorem: real-coefficient polynomials have complex roots in conjugate pairs
Vieta for cubic $az^3 + bz^2 + cz + d$: $\sum\alpha = -b/a$, $\sum\alpha\beta = c/a$, $\alpha\beta\gamma = -d/a$
Conjugate pair $\alpha, \bar\alpha$ gives factor $z^2 - 2\,\text{Re}(\alpha)\,z + |\alpha|^2$

Understand

Concepts

Why the conjugate root theorem fails for complex-coefficient polynomials
Why a degree-$n$ polynomial has exactly $n$ roots (counted with multiplicity) in $\mathbb{C}$
How Vieta turns "one root given" into equations for the remaining roots

Can do

Skills

Solve real-coefficient cubics and quartics given one complex root
Use Vieta to find remaining roots given sum, product or one root
Solve complex-coefficient polynomials by setting up sum/product systems

Key terms

Factor theoremFor a polynomial $P(z)$, $z = \alpha$ is a root iff $(z - \alpha)$ is a factor of $P(z)$. Holds over $\mathbb{C}$ for any coefficients.

Conjugate root theoremIf $P(z)$ has real coefficients and $P(\alpha) = 0$, then $P(\bar\alpha) = 0$. Complex roots occur in conjugate pairs.

Vieta's formulasRelations between the roots and coefficients of a polynomial. For $az^n + \ldots + k$: sum of roots $= -b/a$, product $= (-1)^n k/a$.

Real quadratic factorProduct $(z - \alpha)(z - \bar\alpha) = z^2 - 2\,\text{Re}(\alpha)\,z + |\alpha|^2$, always real, always positive discriminant zero or negative.

MultiplicityThe number of times a root appears as a factor. Counted in the fundamental theorem of algebra: a degree-$n$ polynomial has $n$ roots in $\mathbb{C}$ with multiplicity.

Polynomial long divisionMethod of dividing $P(z)$ by a known factor to find the quotient polynomial whose roots are the remaining roots of $P(z)$.

MEX-N2NESA outcome (Complex Numbers II): uses complex numbers to solve polynomial equations, including applying the conjugate root theorem and relations between roots and coefficients.

Conjugate root theorem and real quadratic factors

core concept

The fundamental theorem of algebra says any degree-$n$ polynomial has exactly $n$ roots in $\mathbb{C}$. When all coefficients of $P(z)$ are real, an additional structure appears: complex roots arrive in conjugate pairs.

$$\text{If } P(\alpha) = 0 \text{ and } P \text{ has real coefficients, then } P(\bar\alpha) = 0.$$

Why? Taking conjugates of $P(\alpha) = 0$ and using $\overline{a_k \alpha^k} = a_k \bar\alpha^k$ (because $\bar{a_k} = a_k$) gives $P(\bar\alpha) = 0$.

Consequence. A conjugate pair $\alpha, \bar\alpha$ produces a real quadratic factor:

$$(z - \alpha)(z - \bar\alpha) = z^2 - (\alpha + \bar\alpha)z + \alpha\bar\alpha = z^2 - 2\,\text{Re}(\alpha)\,z + |\alpha|^2.$$

Worked through the hook: $P(z) = z^3 - 5z^2 + 11z - 15$ has real coefficients and $1 + 2i$ is a root. So $1 - 2i$ is also a root, giving the real quadratic factor $z^2 - 2z + 5$. Dividing: $P(z) = (z^2 - 2z + 5)(z - 3)$, so the third root is $z = 3$. Vieta cross-check: sum $= (1+2i) + (1-2i) + 3 = 5 = -(-5)/1$ ✓.

Warning. The conjugate root theorem requires real coefficients. For polynomials with complex coefficients (e.g., $z^2 - iz + 2 = 0$) conjugate pairs do NOT generally appear, and you must solve by other means.

Real-coefficient $P(z)$: complex roots in conjugate pairs $\alpha, \bar\alpha$ · Conjugate pair factor: $z^2 - 2\,\text{Re}(\alpha)\,z + |\alpha|^2$ · Divide $P(z)$ by this quadratic to reduce degree by 2 · Conjugate root theorem FAILS for complex coefficients

Pause, copy the conjugate-pair quadratic factor $z^2-2\operatorname{Re}(\alpha)z+|\alpha|^2$, the strategy of dividing by it to reduce degree, and the failure of the conjugate-root theorem for complex coefficients into your book.

Quick check: The polynomial $P(z) = z^3 + 2z^2 + 9z + 18$ has real coefficients. If $z = 3i$ is a root, what is the real quadratic factor that $z = 3i$ contributes?

Vieta's formulas: roots from coefficients

core concept

We just saw that real-coefficient polynomials factor into real quadratics $(z^2-2\operatorname{Re}(\alpha)z+|\alpha|^2)$ from each conjugate pair, reducing the degree by 2. That raises a question: now that we can factor, what do Vieta's formulas add for a cubic or quartic? This card answers it → $\sum\alpha = -b$, $\sum\alpha\beta = c$, $\alpha\beta\gamma = -d$ for a monic cubic, with the sign pattern $(-1)^k \cdot (\text{coeff of }z^{n-k})$ for $k$-fold products.

Vieta's formulas express the elementary symmetric functions of the roots in terms of the coefficients. For a monic polynomial they are especially clean.

Cubic. If $\alpha, \beta, \gamma$ are the roots of $z^3 + bz^2 + cz + d = 0$, then:

$$\alpha + \beta + \gamma = -b, \qquad \alpha\beta + \beta\gamma + \gamma\alpha = c, \qquad \alpha\beta\gamma = -d.$$

Quartic. If $\alpha, \beta, \gamma, \delta$ are the roots of $z^4 + bz^3 + cz^2 + dz + e = 0$:

$$\sum \alpha = -b, \qquad \sum \alpha\beta = c, \qquad \sum \alpha\beta\gamma = -d, \qquad \alpha\beta\gamma\delta = e.$$

Why useful? When you know one root or one symmetric function (e.g., "the sum of two roots is 4"), Vieta turns the remaining roots into solutions of a smaller system.

Strategy for "given one root". Use the conjugate root theorem (if real coefficients) to get a second root for free, then either factor out the real quadratic or use Vieta's sum and product to set up two equations in the remaining roots.

Monic cubic $z^3 + bz^2 + cz + d$: $\sum\alpha = -b$, $\sum\alpha\beta = c$, $\alpha\beta\gamma = -d$ · Monic quartic: $\sum\alpha = -b$, $\sum\alpha\beta = c$, $\sum\alpha\beta\gamma = -d$, $\prod\alpha = e$ · Signs alternate: $(-1)^k \cdot (\text{coeff of } z^{n-k}) / (\text{leading coeff})$ for sum of $k$-products · Always check Vieta against the original polynomial as a final verification

Pause, copy Vieta's formulas for monic cubic and quartic (with sign pattern $(-1)^k$), and the verification step (check against original polynomial) into your book.

Did you get this? True or false: if $\alpha, \beta, \gamma$ are roots of $z^3 - 6z^2 + 11z - 6 = 0$, then $\alpha + \beta + \gamma = 6$ and $\alpha\beta\gamma = 6$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · ONE ROOT GIVEN, REAL COEFFICIENTS

Given that $z = 2 - i$ is a root of $P(z) = z^3 - 7z^2 + 17z - 15$, find all roots.

Coefficients of $P(z)$ are real, so by the conjugate root theorem $z = 2 + i$ is also a root. The corresponding real quadratic factor is $(z - (2-i))(z - (2+i)) = z^2 - 4z + 5$.

Always check coefficients are real before applying the conjugate root theorem. Compute $\alpha + \bar\alpha = 4$ and $\alpha\bar\alpha = |\alpha|^2 = 5$ to get the real quadratic.

PROBLEM 2 · VIETA WITH PARTIAL INFORMATION

The cubic $z^3 - 4z^2 + kz - 13 = 0$ has real coefficients. If one root is $z = 2 + 3i$, find $k$ and the third root.

By the conjugate root theorem, $z = 2 - 3i$ is also a root. Let the third root be $\gamma$ (real). Use Vieta's sum: $(2+3i) + (2-3i) + \gamma = -(-4) = 4$, so $4 + \gamma = 4$, giving $\gamma = 0$.

When one root is complex with non-zero imaginary part, the imaginary parts cancel in the sum, leaving just the third (real) root to solve for.

PROBLEM 3 · QUARTIC WITH TWO COMPLEX PAIRS

Solve $z^4 - 2z^3 + 6z^2 - 2z + 5 = 0$, given that $z = i$ is a root.

Coefficients are real, so $z = -i$ is also a root. Real quadratic factor: $(z - i)(z + i) = z^2 + 1$.

For purely imaginary roots $\pm i\beta$ the real quadratic factor is $z^2 + \beta^2$, the linear term vanishes because $\text{Re}(\alpha) = 0$.

Fill the gap: If $z = 1 + i\sqrt{3}$ is a root of a real-coefficient polynomial, the conjugate root is , and these two roots together contribute the real quadratic factor $z^2 - \;$$\,z + \;$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Applying conjugate root theorem to complex coefficients

The conjugate root theorem requires real coefficients. For $z^2 - iz + 2 = 0$ the conjugates of roots are NOT also roots. Always inspect the coefficients before assuming conjugate pairs.

Trap 02

Sign errors in Vieta

Vieta alternates signs: $\sum\alpha = -b/a$ (negative), $\sum\alpha\beta = c/a$ (positive), $\alpha\beta\gamma = -d/a$ (negative). Forgetting the alternating sign is the most common Vieta error, write the rule down before substituting.

Trap 03

Multiplying conjugate pair carelessly

$(z - \alpha)(z - \bar\alpha) = z^2 - 2\,\text{Re}(\alpha)\,z + |\alpha|^2$, the linear coefficient is $-2\,\text{Re}(\alpha)$ (not $-\text{Re}(\alpha)$) and the constant is $|\alpha|^2$ (not $\alpha^2$). Use this formula directly; do not expand from scratch each time.

Did you get this? True or false: the polynomial $z^2 - (3+i)z + (2 + 2i) = 0$ has $z = 1 - i$ as a root because $1 - i$ is the conjugate of $1 + i$, which is a root.

Activities · practice with the ideas

Given that $z = 2i$ is a root of $z^3 - 3z^2 + 4z - 12 = 0$, find all roots and confirm using Vieta's sum.

Find a monic cubic with real coefficients whose roots are $z = 1 - i$ and $z = 4$. Write out the expansion.

The cubic $z^3 + pz^2 + qz + r = 0$ has roots $\alpha, \beta, \gamma$ with $\alpha + \beta = 3$ and $\gamma = 2$. If $\alpha\beta = 5$, find $p, q, r$.

Solve $z^4 + 4 = 0$ over $\mathbb{C}$. (Hint: factor as a difference of squares using complex numbers, or write $z^4 = -4 = 4e^{i\pi}$.)

The cubic $z^3 + (1-2i)z^2 - (3+2i)z + 3i = 0$ has complex coefficients. Verify $z = i$ is a root, then factor and find the remaining roots.

Odd one out: Three of these statements about a real-coefficient cubic with roots $\alpha, \beta, \gamma$ are correct. Which one is NOT?

Revisit your thinking

Earlier you predicted (a) another root of $z^3 - 5z^2 + 11z - 15$ given that $1 + 2i$ is a root, and (b) the sum of all three roots.

The other obvious root is the conjugate $z = 1 - 2i$, because the polynomial has real coefficients. The sum of all three roots equals $-(-5)/1 = 5$ by Vieta, which means the third (real) root must be $5 - (1+2i) - (1-2i) = 3$. Checking: $(z - 3)$ should divide $P(z)$, and indeed $P(3) = 27 - 45 + 33 - 15 = 0$ ✓. The three-tool combination (conjugate, factor, Vieta) is the standard HSC strategy.

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Multiple choice

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Short answer

ApplyBand 32 marks

Q1. Given that $z = 3 + i$ is a root of $z^3 - 8z^2 + 26z - 30 = 0$, find the other two roots. (2 marks)

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ApplyBand 43 marks

Q2. The quartic $z^4 - 4z^3 + 14z^2 - 36z + 45 = 0$ has $z = 1 + 2i$ as a root. Find all four roots. (3 marks)

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AnalyseBand 53 marks

Q3. The cubic $z^3 + bz^2 + cz + d = 0$ has real coefficients and roots $\alpha, \beta, \gamma$ satisfying $\alpha + \beta + \gamma = -2$, $\alpha\beta + \beta\gamma + \gamma\alpha = 5$, and one root is $\gamma = -3$. Find the cubic and the two remaining roots. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Real coefficients $\Rightarrow z = -2i$ also a root. Factor: $z^2 + 4$. Divide: $z^3 - 3z^2 + 4z - 12 = (z^2 + 4)(z - 3)$. Roots: $2i, -2i, 3$. Vieta sum: $0 + 3 = 3 = -(-3)$ ✓.

2. Conjugate of $1 - i$ is $1 + i$. Quadratic: $z^2 - 2z + 2$. With linear factor $z - 4$: $(z^2 - 2z + 2)(z - 4) = z^3 - 6z^2 + 10z - 8$.

3. $p = -(\alpha + \beta + \gamma) = -5$. $q = \alpha\beta + \gamma(\alpha + \beta) = 5 + 2 \cdot 3 = 11$. $r = -\alpha\beta\gamma = -5 \cdot 2 = -10$.

4. $z^4 = -4 = 4e^{i\pi}$. $z = \sqrt{2}\,e^{i(\pi + 2k\pi)/4}$, $k = 0,1,2,3$. Roots: $1+i, -1+i, -1-i, 1-i$ (two conjugate pairs).

5. Test $z = i$: $i^3 + (1-2i)i^2 - (3+2i)i + 3i$. Compute term-by-term and divide; the remaining quadratic factor yields the other two roots via the quadratic formula.

Q1 (2 marks): Conjugate $3 - i$ is also a root [1]. Quadratic factor $z^2 - 6z + 10$; dividing gives third linear factor $z - 3$, so third root is $z = 3$ [1].

Q2 (3 marks): Conjugate $1 - 2i$ also a root [1]. Quadratic factor $z^2 - 2z + 5$; dividing into the quartic gives quotient $z^2 - 2z + 9$ [1]. Solve quotient: $z = 1 \pm 2i\sqrt{2}$ [1]. Four roots: $1 + 2i,\; 1 - 2i,\; 1 + 2i\sqrt{2},\; 1 - 2i\sqrt{2}$.

Q3 (3 marks): From Vieta: $b = 2$, $c = 5$ [1]. With $\gamma = -3$: $\alpha + \beta = 1$ and $\alpha\beta = 8$ (from $\alpha\beta + \gamma(\alpha+\beta) = 5$) [1]. So $\alpha, \beta$ are roots of $z^2 - z + 8 = 0$, giving $z = \tfrac{1 \pm i\sqrt{31}}{2}$, and $d = -\alpha\beta\gamma = 24$ [1]. Cubic: $z^3 + 2z^2 + 5z + 24 = 0$.

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Module overview · Extension 2