Complex Polynomials and the Fundamental Theorem of Algebra
The fundamental theorem of algebra states that every non-constant polynomial with complex coefficients has at least one complex root, and therefore a polynomial of degree n has exactly n complex roots counted with multiplicity. In brief: z3 − 1 has three complex roots, 1 and (−1 ± √3 i)/2, and every polynomial factors completely over the complex numbers as p(z) = a(z − z1)(z − z2)…(z − zn). For real coefficients, nonreal roots come in conjugate pairs.
Over the real numbers, x2 + 1 has no root, and polynomials of higher degree may have fewer than n real roots. Over the complex numbers, the situation is perfect: the number of roots always equals the degree. This is the main reason why complex numbers are the natural setting for algebra. This guide states the theorem, gives its consequences for factoring, explains the conjugate root theorem and the factorization of real polynomials into linear and quadratic factors, with examples and a table.
What does the fundamental theorem say?
Every polynomial p(z) = anzn + … + a1z + a0 of degree n ≥ 1 with complex coefficients has at least one complex root. The name is traditional: the theorem is really about analysis, and its proofs use ideas from complex analysis or topology. The theorem holds for real coefficients too, since real numbers are complex numbers.
The theorem guarantees one root. With the factor theorem one can divide out (z − z1) to get a polynomial of degree n − 1, and repeat. Therefore p(z) = an(z − z1)(z − z2)…(z − zn), with roots listed according to multiplicity.
What are the consequences?
- A polynomial of degree n has exactly n complex roots counted with multiplicity.
- Every polynomial factors completely into linear factors over ℂ. There are no irreducible polynomials of degree greater than 1 over the complex numbers.
- Vieta’s formulas relate the roots to the coefficients: for z2 + bz + c the sum of the roots is −b and the product is c. For degree n, the sum of the roots is −an−1/an and the product is (−1)na0/an.
- The roots depend continuously on the coefficients.
Example 1: counting
p(z) = z4 + 4 has 4 roots. By the formula for roots of complex numbers, z4 = −4 gives z = √2(cos(45° + 90°k) + i sin(45° + 90°k)) = ±1 ± i. So z4 + 4 = (z − 1 − i)(z − 1 + i)(z + 1 − i)(z + 1 + i) = (z2 − 2z + 2)(z2 + 2z + 2).
What is the conjugate root theorem?
If p has real coefficients and z0 = a + bi is a root, then so is the conjugate z̄0 = a − bi. The reason: p(z̄0) = conj(p(z0)) = conj(0) = 0, since conjugation respects sums and products and leaves real coefficients unchanged. So nonreal roots of real polynomials come in pairs, and a real polynomial of odd degree has at least one real root.
The pair (z − (a + bi))(z − (a − bi)) = z2 − 2az + (a2 + b2) is a real quadratic with a negative discriminant.
How do real polynomials factor?
Every real polynomial factors over ℝ into linear factors (from real roots) and irreducible quadratic factors (from pairs of conjugate nonreal roots). Examples: x4 + 4 = (x2 − 2x + 2)(x2 + 2x + 2); x3 − 1 = (x − 1)(x2 + x + 1); x4 + 1 = (x2 + √2 x + 1)(x2 − √2 x + 1).
Worked examples
Example 2: finding the other roots
One root of z3 − 3z2 + 4z − 2 = 0 is 1 (1 − 3 + 4 − 2 = 0). Divide by (z − 1): z2 − 2z + 2. The roots of the quadratic: z = (2 ± √(4 − 8))/2 = 1 ± i. So the roots are 1, 1 + i and 1 − i. A conjugate pair appears, as the coefficients are real.
Example 3: building a polynomial from its roots
A real polynomial of degree 4 has the roots 2 and 3 − i and the leading coefficient 1, and 2 is a double root. Then the roots are 2, 2, 3 − i and 3 + i (the conjugate). The polynomial is (z − 2)2[z2 − 6z + 10] = (z2 − 4z + 4)(z2 − 6z + 10) = z4 − 10z3 + 38z2 − 64z + 40. Check at z = 2: 16 − 80 + 152 − 128 + 40 = 0 ✓.
Example 4: a polynomial with complex coefficients
z2 − 2iz − 5 = 0. D = (−2i)2 + 20 = −4 + 20 = 16. z = (2i ± 4)/2 = 2 + i or −2 + i. The roots are not conjugates, as the coefficients are not real. Check by Vieta: sum = 2i ✓ and product = (2 + i)(−2 + i) = −4 + 2i − 2i + i2 = −5 ✓.
Example 5: sum and product of roots
For z3 − 6z2 + 11z − 6 = 0, the sum of the roots is 6 and the product is 6. The roots are 1, 2, 3 ✓. For z4 + 4 the sum of the roots is 0 and the product is 4: (1 + i)(1 − i)(−1 + i)(−1 − i) = 2 · 2 = 4 ✓.
Example 6: a real polynomial of odd degree
x5 + x + 1 has degree 5. Since non-real roots come in pairs, there are 1, 3 or 5 real roots. It has exactly one real root, near −0.755 (since the function is increasing: the derivative 5x4 + 1 > 0), and four nonreal roots forming two conjugate pairs.
Table of roots of sample polynomials
| Polynomial | Degree | Roots | Real roots |
|---|---|---|---|
| z2 + 1 | 2 | i, −i | 0 |
| z2 − 2z + 5 | 2 | 1 + 2i, 1 − 2i | 0 |
| z3 − 1 | 3 | 1, (−1 + √3 i)/2, (−1 − √3 i)/2 | 1 |
| z3 − 3z2 + 4z − 2 | 3 | 1, 1 + i, 1 − i | 1 |
| z4 + 4 | 4 | 1 + i, 1 − i, −1 + i, −1 − i | 0 |
| (z − 1)2(z + 2) | 3 | 1 (double), −2 | 3 counting multiplicity |
Where does the theorem come from?
Albert Girard stated in 1629 that an equation of degree n has n roots, and d’Alembert gave an attempted proof in 1746, which was incomplete. Gauss gave the first widely accepted proof in his doctoral thesis in 1799 and later gave several more. Modern proofs use Liouville’s theorem from complex analysis or algebraic topology. There is no algebraic formula for the roots of a general polynomial of degree 5 or higher, as proved by Abel and Ruffini, but the theorem guarantees that the roots exist.
Where is it used today?
In control engineering, the stability of a system depends on the location of the roots of the characteristic polynomial in the complex plane: if all have a negative real part, the system is stable. In linear algebra, the eigenvalues of a matrix are the roots of its characteristic polynomial, and the theorem guarantees that every square matrix has complex eigenvalues. Numerical software finds roots of polynomials of high degree by iteration.
Which mistakes are common?
Forgetting multiplicity
A double root counts twice. The polynomial (z − 1)2 has two roots (both equal to 1).
Assuming that real polynomials have only real roots
x2 + 1 has two complex roots and no real ones.
Applying the conjugate root theorem to complex coefficients
The pairing holds only if all coefficients are real. For z2 − 2iz − 5, the roots 2 + i and −2 + i are not conjugates.
Thinking the theorem gives a formula for the roots
It asserts the existence of the roots, not an algorithm. Formulas exist for degrees up to 4.