Polynomial Equations with Complex Roots
A polynomial of degree $n$ has exactly $n$ roots in $\mathbb{C}$ — counted with multiplicity. That single fact, the Fundamental Theorem of Algebra, transforms how we read every polynomial. This lesson combines $z^n = c$, the factor theorem extended to complex factors, and quadratics with complex coefficients into one toolbox for solving any low-degree polynomial in $\mathbb{C}$.
The polynomial $P(z) = z^2 + 1$ has no real roots. Before checking — write down its two complex roots, then factor $P(z)$ over $\mathbb{C}$ as a product of linear factors. Can you do the same for $z^3 - 1$? Sketch your reasoning.
Every polynomial $P(z)$ equation rewards two habits: count the roots before solving ($\deg P = n$ means $n$ roots in $\mathbb{C}$ counted with multiplicity), then factor strategically — extract any roots you can spot, use the factor theorem to peel off linear factors, then attack the remaining quadratic with the quadratic formula. The quadratic formula still works when coefficients are complex.
The count-spot-peel reading: (1) note the degree $n$ — there are $n$ roots; (2) spot or test small roots ($\pm 1, \pm i, \dots$); (3) divide by $(z - r)$ to reduce the degree; repeat or finish with the quadratic formula in $\mathbb{C}$.
$z = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ · works for $a, b, c \in \mathbb{C}$
Key facts
- FTA: every degree-$n$ polynomial has exactly $n$ roots in $\mathbb{C}$ (with multiplicity)
- Factor theorem: $z = r$ is a root $\iff (z - r)$ divides $P(z)$
- Quadratic formula works over $\mathbb{C}$ — discriminant can be any complex number
- Real-coefficient polynomials: non-real roots come in conjugate pairs
Concepts
- Why a polynomial over $\mathbb{C}$ always factors into linear factors
- How $z^n = c$ is the simplest polynomial equation and seeds the rest
- Why complex coefficients break the conjugate-pair rule
Skills
- Solve $z^n = c$ for small $n$ and factor as $(z - z_1)\cdots(z - z_n)$
- Use a known root to reduce a cubic/quartic via the factor theorem
- Solve $z^2 + bz + c = 0$ for $b, c \in \mathbb{C}$ using the quadratic formula
The Fundamental Theorem of Algebra (FTA) guarantees that any non-constant polynomial $P(z)$ has at least one complex root. Combined with the factor theorem, this means a degree-$n$ polynomial factors completely into linear factors over $\mathbb{C}$:
where $z_1, z_2, \dots, z_n$ are the roots of $P$ in $\mathbb{C}$ (some may coincide — multiplicity). This is the structural reason every polynomial equation $P(z) = 0$ has exactly $\deg P$ solutions.
Worked through the hook: $z^2 + 2z + 5 = 0$. Discriminant $\Delta = 4 - 20 = -16$, so $\sqrt{\Delta} = \pm 4i$ and
$$z = \frac{-2 \pm 4i}{2} = -1 \pm 2i.$$
Both roots non-real, conjugate pair (because the coefficients are real). The polynomial factors as $z^2 + 2z + 5 = (z - (-1+2i))(z - (-1-2i))$. Two roots — exactly the degree.
FTA: $\deg P = n \Rightarrow n$ roots in $\mathbb{C}$ (with multiplicity) · Factor theorem: $P(r) = 0 \iff (z - r) \mid P(z)$ · Real coefficients $\Rightarrow$ non-real roots come in conjugate pairs · Strategy: count, spot, peel, finish with quadratic formula
Pause — copy the FTA statement ($n$ roots with multiplicity), the factor theorem, the conjugate-pair rule (real coefficients only), and the strategy (count, spot, peel, quadratic) into your book.
Quick check: The polynomial $P(z) = z^4 - 1$ factors completely over $\mathbb{C}$ as which of the following?
We just saw that the Fundamental Theorem of Algebra guarantees $n$ roots in $\mathbb{C}$ for a degree-$n$ polynomial, and that real coefficients force non-real roots into conjugate pairs. That raises a question: does the quadratic formula still work over $\mathbb{C}$ when coefficients are complex? This card answers it → yes, but you must find $\sqrt{\Delta}$ in $\mathbb{C}$ using the algebraic method, and the conjugate-root theorem no longer applies.
The quadratic formula $z = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ remains valid when $a, b, c$ are any complex numbers — but two things change:
- The discriminant $\Delta = b^2 - 4ac$ is itself a complex number. Use the techniques of Lesson 03 to find both square roots of $\Delta$.
- The conjugate-root theorem fails. When a coefficient is non-real, the two roots are generally unrelated complex numbers — not conjugates of each other.
Quadratic formula works over $\mathbb{C}$ for any $a, b, c$ · Compute $\Delta = b^2 - 4ac$; find both square roots of $\Delta$ in $\mathbb{C}$ · Conjugate-root theorem requires real coefficients — otherwise it fails · For complex $\Delta$, use the algebraic ($a^2 - b^2$, $2ab$) method to extract $\sqrt{\Delta}$
Pause — copy the steps for complex-coefficient quadratics: compute $\Delta = b^2-4ac$ in $\mathbb{C}$, find both square roots of $\Delta$, apply the formula; note the conjugate-root theorem fails here into your book.
Did you get this? True or false: if $P(z) = z^3 + z + (1 - i)$, then knowing one root automatically tells you another root via the conjugate-root theorem.
Worked examples · 3 in a row, reveal as you go
Find all four roots of $z^4 = -4$ and hence factor $z^4 + 4$ as a product of linear factors over $\mathbb{C}$.
Given that $z = 1 + i$ is a root of $P(z) = z^3 - 3z^2 + 4z - 2$, find all three roots.
Solve $z^2 - (3 + i)z + (2 + 2i) = 0$ over $\mathbb{C}$.
Fill the gap: The Fundamental Theorem of Algebra states that any polynomial of degree with complex coefficients has exactly roots in $\mathbb{C}$, counted with multiplicity.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the polynomial $(z - 1)^2(z^2 + 4)$ has four roots in $\mathbb{C}$ when counted with multiplicity.
Activities · practice with the ideas
Factor $z^3 - 1$ as a product of linear factors over $\mathbb{C}$.
Given that $z = 2 - i$ is a root of $P(z) = z^3 - 5z^2 + 9z - 5$, find the other two roots.
Solve $z^2 + iz + 2 = 0$ over $\mathbb{C}$. Are the roots conjugates of each other?
Factor $z^6 - 1$ over $\mathbb{C}$ as a product of six linear factors.
Construct a real-coefficient quadratic with roots $3 + 2i$ and $3 - 2i$. Verify by expanding.
Odd one out: Three of these statements are true for every polynomial $P(z)$ of degree $n \geq 1$ with complex coefficients. Which one is NOT?
Earlier you solved $z^2 + 2z + 5 = 0$, then tried $z^2 + (1-i)z - i = 0$.
The first has real coefficients: roots $-1 \pm 2i$, a conjugate pair, exactly as the conjugate-root theorem predicts. The second has a non-real coefficient ($1 - i$): factoring as $(z + 1)(z - i)$ gives roots $z = -1$ and $z = i$ — completely unrelated. Same quadratic formula, same FTA-guaranteed two roots, but the conjugate symmetry is gone. The lesson: count first (FTA tells you how many), then check coefficients (real or complex) to decide whether conjugate-pair shortcuts apply.
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. Factor $z^4 - 16$ as a product of linear factors over $\mathbb{C}$. (2 marks)
Q2. Given that $z = 1 + 2i$ is a root of $P(z) = z^3 - 7z^2 + 17z - 15$, find the other two roots. (3 marks)
Q3. Solve $z^2 - (2 + 2i)z + (3i) = 0$ over $\mathbb{C}$. State whether the two roots are conjugates and justify. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $z^3 = 1$: roots $1$, $\omega = \text{cis}(2\pi/3) = \tfrac{-1 + \sqrt{3}\,i}{2}$, $\omega^2 = \text{cis}(4\pi/3) = \tfrac{-1 - \sqrt{3}\,i}{2}$. So $z^3 - 1 = (z - 1)(z - \omega)(z - \omega^2)$.
2. $P(z) = z^3 - 5z^2 + 9z - 5$. Real coefficients $\Rightarrow$ $2 + i$ is also a root. Quadratic factor: $(z - (2-i))(z - (2+i)) = z^2 - 4z + 5$. Divide: $P(z) = (z^2 - 4z + 5)(z - 1)$. Third root $z = 1$.
3. $z^2 + iz + 2 = 0$. $\Delta = i^2 - 8 = -1 - 8 = -9$. $\sqrt{-9} = \pm 3i$. $z = \tfrac{-i \pm 3i}{2}$: roots $z = i$ and $z = -2i$. Not conjugates because $b = i$ is non-real.
4. $z^6 = 1$: roots $\text{cis}(\pi k / 3)$ for $k = 0, \dots, 5$ — i.e. $1,\;\tfrac{1+\sqrt{3}\,i}{2},\;\tfrac{-1+\sqrt{3}\,i}{2},\;-1,\;\tfrac{-1-\sqrt{3}\,i}{2},\;\tfrac{1-\sqrt{3}\,i}{2}$. $z^6 - 1$ factors as the product of $(z - r)$ over these six roots.
5. Sum of roots $= 6$, product $= (3)^2 - (2i)^2 = 9 + 4 = 13$. So quadratic is $z^2 - 6z + 13$. Verify: $(z-3)^2 + 4 = z^2 - 6z + 9 + 4 = z^2 - 6z + 13$ ✓.
Q1 (2 marks): $z^4 = 16$, so $z = 2\,\text{cis}(\pi k/2)$ for $k = 0,1,2,3$, giving $z = 2, 2i, -2, -2i$ [1]. Therefore $z^4 - 16 = (z - 2)(z + 2)(z - 2i)(z + 2i)$ [1].
Q2 (3 marks): $P$ has real coefficients, so $\overline{1 + 2i} = 1 - 2i$ is also a root [1]. Quadratic factor $(z - (1+2i))(z - (1-2i)) = z^2 - 2z + 5$ [1]. Dividing $P(z)$ by $z^2 - 2z + 5$ yields quotient $z - 3$, so the third root is $z = 3$ [1].
Q3 (3 marks): $\Delta = (2 + 2i)^2 - 4 \cdot 3i = (4 + 8i - 4) - 12i = -4i$ [1]. Find $\sqrt{-4i}$: set $(p + qi)^2 = -4i$ giving $p^2 - q^2 = 0$, $2pq = -4$, so $p = -q$ and $pq = -2$, hence $\sqrt{-4i} = \pm(\sqrt{2} - \sqrt{2}\,i)$ [1]. Then $z = \dfrac{(2 + 2i) \pm (\sqrt{2} - \sqrt{2}\,i)}{2}$, giving $z = 1 + \tfrac{\sqrt{2}}{2} + (1 - \tfrac{\sqrt{2}}{2})i$ or $z = 1 - \tfrac{\sqrt{2}}{2} + (1 + \tfrac{\sqrt{2}}{2})i$. Not conjugates, because the coefficients of the quadratic are not all real [1].
Five timed questions on factoring polynomials over $\mathbb{C}$, the factor theorem and quadratics with complex coefficients. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick polynomial-roots questions. Lighter alternative to the boss.
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