Every polynomial equation with complex coefficients has at least one root in the complex number field

Every polynomial equation with complex coefficients has at least one root in the complex number field. This fundamental result is known as the Fundamental Theorem of Algebra. It states that every non-constant polynomial equation of degree greater than or equal to 1 has at least one complex root.

The complex number field, denoted as ℂ, consists of all numbers of the form a + bi, where “a” and “b” are real numbers, and “i” represents the imaginary unit (i.e., √(-1)). In other words, the complex numbers extend the real numbers by adding an additional dimension, the imaginary part.

The Fundamental Theorem of Algebra can be proven using various methods, but one common approach is to use contradiction. Suppose you have a polynomial equation of degree “n” with complex coefficients, and you assume it has no complex roots. Then, you can construct a new polynomial equation of degree “n” whose roots are precisely the reciprocals (or negatives) of the original polynomial’s roots. This new polynomial will also have complex coefficients and no complex roots. However, by repeatedly applying this process, you eventually reach a contradiction because you would obtain a constant polynomial with no complex roots, which is impossible.

Therefore, the existence of at least one complex root for any non-constant polynomial equation is guaranteed by the Fundamental Theorem of Algebra. This property makes the complex numbers a crucial extension of the real numbers and plays a fundamental role in many areas of mathematics and science.

The proof of the Fundamental Theorem of Algebra is a classical and fundamental result in complex analysis and algebra. There are several ways to prove it, and I’ll provide a detailed outline of one common proof. This proof relies on the properties of complex numbers and complex analysis.

Theorem: (Fundamental Theorem of Algebra) Every non-constant polynomial equation with complex coefficients has at least one complex root.

Proof:

Assume the Contrary: Let’s begin by assuming the opposite, that there exists a non-constant polynomial equation with complex coefficients that has no complex roots.
Degree Reduction: Consider a polynomial equation of degree “n” with complex coefficients: P(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀ We can assume that the leading coefficient aₙ is nonzero, as otherwise, we could reduce the degree until we reach a nonzero coefficient.
Reciprocal Roots: Since we’re assuming there are no complex roots, we can consider its reciprocal polynomial: Q(z) = 1/P(z) = 1/(aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀) If P(z) has no complex roots, then Q(z) is well-defined for all complex numbers z.
Degree Preservation: The degree of Q(z) is the same as the degree of P(z), which is “n.” This is because the reciprocal of a polynomial of degree “n” will still have a leading term of degree “n.”
Bounding Q(z): Consider the modulus (absolute value) of Q(z) for a complex number z ≠ 0: |Q(z)| = 1/|P(z)| Since |P(z)| is always positive (because its coefficients are complex, not zero), |Q(z)| is also bounded and non-zero for all complex numbers.
Liouville’s Theorem (Complex Analysis): Liouville’s theorem states that if a bounded entire function (a function that’s holomorphic on the entire complex plane and bounded) is constant. In this case, Q(z) is an entire function that is bounded for all complex numbers, which implies it is constant.
Constant Function: Since Q(z) is constant, it means that 1/P(z) is constant, which implies that P(z) is also constant.
Contradiction: However, we started with a non-constant polynomial equation, and we’ve reached the conclusion that P(z) must be constant, which contradicts our initial assumption.
Conclusion: Therefore, our assumption that there exists a non-constant polynomial equation with complex coefficients that has no complex roots must be false. Hence, every non-constant polynomial equation with complex coefficients has at least one complex root.

This completes the proof of the Fundamental Theorem of Algebra. It relies on the principles of complex analysis and the properties of complex numbers to establish the existence of complex roots for non-constant polynomial equations.

Proof 2: Using Compactness in the Complex Plane

Assume the Contrary: Start by assuming that there exists a non-constant polynomial equation with complex coefficients that has no complex roots.
Degree and Leading Coefficient: Consider a polynomial equation of degree “n” with complex coefficients: P(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀ We can assume aₙ ≠ 0 (leading coefficient is nonzero) and that it has no complex roots as per our assumption.
Homogenization: To simplify the argument, we can homogenize the polynomial by dividing through by aₙ: Q(z) = zⁿ + (aₙ₋₁/aₙ)zⁿ⁻¹ + … + (a₀/aₙ) This polynomial is still non-constant and has no complex roots.
Compactness of Closed Discs: Consider the closed disc D(0, R) of radius R centered at the origin in the complex plane. We want to show that Q(z) ≠ 0 for z in this disc.
Continuous Function: Observe that Q(z) is a continuous function on the closed disc, as it is a polynomial with continuous coefficients.
Minimum Value: Since the closed disc is compact, Q(z) attains its minimum value M inside the disc.
Bound Away from Zero: Because we assumed that Q(z) ≠ 0 for any complex number, the minimum value M must be strictly greater than 0. In other words, |Q(z)| > 0 for all z in D(0, R).
Contradiction: However, by the Extreme Value Theorem, a continuous function on a compact set must attain its minimum and maximum values. This means that |Q(z)| attains a minimum M within the disc, but we just showed that |Q(z)| is strictly greater than 0 for all z in the disc. This is a contradiction.
Conclusion: The contradiction arises from our initial assumption that there exists a non-constant polynomial equation with complex coefficients that has no complex roots. Therefore, this assumption must be false.

Proof 3: Using Limits and Bolzano-Weierstrass Theorem

Assume the Contrary: Assume there exists a non-constant polynomial equation with complex coefficients that has no complex roots.
Degree and Leading Coefficient: Consider a polynomial equation of degree “n” with complex coefficients: P(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀ Assume aₙ ≠ 0 and that it has no complex roots as per our assumption.
Scaling: Define a new polynomial by scaling P(z) as follows: Q(z) = P(z) / (aₙzⁿ) This new polynomial has the same degree as P(z) but has no constant term (it’s a monic polynomial). Therefore, Q(z) ≠ 0 for all complex numbers.
Behavior at Infinity: Consider the behavior of Q(z) as |z| approaches infinity. Since Q(z) ≠ 0, we have: lim |z| → ∞ Q(z) ≠ 0
Polynomial Limit: As |z| approaches infinity, all the higher-degree terms in Q(z) become negligible compared to the leading term zⁿ. Therefore: lim |z| → ∞ Q(z) = lim |z| → ∞ (zⁿ / (aₙzⁿ))
Simplify the Limit: The limit can be simplified to: lim |z| → ∞ (zⁿ / (aₙzⁿ)) = 1/aₙ
Nonzero Limit: Since the limit is a nonzero complex number (1/aₙ), this means that Q(z) does not tend to 0 as |z| goes to infinity.
Contradiction: However, by the definition of limits, for any ε > 0, there exists a positive number R such that if |z| > R, then |Q(z) – (1/aₙ)| < ε. This means that for sufficiently large |z|, Q(z) is arbitrarily close to 1/aₙ, which contradicts the fact that Q(z) does not tend to 0.
Conclusion: The contradiction arises from our initial assumption that there exists a non-constant polynomial equation with complex coefficients that has no complex roots. Therefore, this assumption must be false.

These two additional proofs illustrate different ways to approach the Fundamental Theorem of Algebra, one using compactness in the complex plane and the other using limits and the Bolzano-Weierstrass Theorem. Each proof provides a rigorous demonstration of why every non-constant polynomial equation with complex coefficients must have at least one complex root.

There are topological proofs of the Fundamental Theorem of Algebra as well. These proofs leverage concepts from topology, particularly the theory of the complex plane, to demonstrate that every non-constant polynomial equation with complex coefficients must have at least one complex root. One notable topological approach is based on the argument principle and Rouche’s theorem. Here’s an outline of this topological proof:

Proof 4: Using the Argument Principle and Rouche’s Theorem

Assume the Contrary: Start by assuming that there exists a non-constant polynomial equation with complex coefficients that has no complex roots.
Degree and Leading Coefficient: Consider a polynomial equation of degree “n” with complex coefficients: P(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀ Assume aₙ ≠ 0 and that it has no complex roots as per our assumption.
Contour Integration: Choose a sufficiently large positive number R and consider the closed contour C_R, which is a circle centered at the origin with radius R.
Argument Principle: Apply the Argument Principle, which states that for a meromorphic function f(z) with no poles on or inside a closed contour C, the difference between the total count of zeros and poles (with multiplicities) enclosed by C is given by: N₀ – P₀ = (1/2πi) ∮_C (f'(z)/f(z)) dz Here, N₀ represents the number of zeros inside C, and P₀ represents the number of poles inside C.
Apply to P(z): Apply the Argument Principle to the polynomial P(z) over the contour C_R. Since P(z) has no complex roots, it has no zeros inside C_R.
No Poles Inside: Additionally, since P(z) is a polynomial (and hence, an entire function), it has no poles anywhere in the complex plane.
Result of Argument Principle: Therefore, according to the Argument Principle, we have: N₀ – P₀ = 0 This implies that there are no zeros of P(z) inside C_R.
Apply Rouche’s Theorem: Now, consider the polynomial P(z) + ε, where ε is a small positive constant. Since P(z) has no zeros inside C_R, the zeros of P(z) + ε will also be outside C_R.
Rouche’s Theorem: Apply Rouche’s Theorem, which states that if two complex functions have the same number of zeros (counted with multiplicities) inside a closed contour and one of the functions dominates the other on the contour, then they have the same number of zeros inside the contour.
Dominance on C_R: Because ε is small, P(z) + ε dominates P(z) on C_R, meaning that |P(z) + ε| > |P(z)| for all z on C_R.
Conclusion: Since P(z) + ε has no zeros inside C_R and dominates P(z) on C_R, it follows that P(z) + ε has no zeros inside C_R either.
Contradiction: However, as R approaches infinity, P(z) + ε becomes a constant nonzero function, which clearly has no zeros. This contradicts the previous conclusion that P(z) + ε has no zeros inside C_R.
Final Conclusion: The contradiction arises from our initial assumption that there exists a non-constant polynomial equation with complex coefficients that has no complex roots. Therefore, this assumption must be false.

This topological proof demonstrates the existence of complex roots for non-constant polynomial equations by utilizing the Argument Principle and Rouche’s Theorem within the framework of complex plane topology.

The idea of treating a polynomial as a map from the complex plane into itself is a topological approach known as the Gauss-Lucas theorem. This theorem utilizes the concept of fixed points in a continuous map to prove the Fundamental Theorem of Algebra. Here’s a proof based on this concept:

Proof 5: Using the Gauss-Lucas Theorem

Assume the Contrary: Start by assuming that there exists a non-constant polynomial equation with complex coefficients that has no complex roots.
Degree and Leading Coefficient: Consider a polynomial equation of degree “n” with complex coefficients: P(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀ Assume aₙ ≠ 0 and that it has no complex roots as per our assumption.
Complex Plane as the Domain: View the complex plane ℂ as the domain, and consider the polynomial P(z) as a map from ℂ to ℂ. P: ℂ → ℂ This map associates each complex number z with another complex number P(z).
Continuous Map: The map P is continuous since it’s a polynomial, and polynomials are continuous functions.
Compactness of the Domain: The complex plane ℂ is a compact space, meaning it is closed and bounded.
Brouwer Fixed-Point Theorem: Apply Brouwer’s Fixed-Point Theorem, which states that any continuous map from a compact convex subset of a Euclidean space to itself has at least one fixed point. In our case, ℂ is a compact convex subset of ℂ.
Fixed Point: By the Brouwer Fixed-Point Theorem, the polynomial map P: ℂ → ℂ must have at least one fixed point, i.e., there exists a complex number z₀ such that P(z₀) = z₀.
Contradiction: However, if P(z₀) = z₀, then z₀ is a complex root of the polynomial equation P(z) = 0. This contradicts our initial assumption that the polynomial has no complex roots.
Conclusion: The contradiction arises from our initial assumption that there exists a non-constant polynomial equation with complex coefficients that has no complex roots. Therefore, this assumption must be false.

This proof employs the concept of fixed points in a continuous map from the complex plane to itself, relying on the Brouwer Fixed-Point Theorem to establish the existence of a complex root for the polynomial.

Certainly, it’s possible to prove the Fundamental Theorem of Algebra without explicitly invoking the concept of fixed points. One such approach involves using the winding number of paths in the complex plane to demonstrate that a polynomial with no complex roots would lead to contradictions. Here’s an outline of this proof:

Proof 6: Using Winding Numbers

Assume the Contrary: Start by assuming that there exists a non-constant polynomial equation with complex coefficients that has no complex roots.
Degree and Leading Coefficient: Consider a polynomial equation of degree “n” with complex coefficients: P(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀ Assume aₙ ≠ 0 and that it has no complex roots as per our assumption.
Complex Plane Paths: Now, consider any closed path γ in the complex plane, which encircles the origin (i.e., γ is a closed curve that loops around the origin).
Winding Number: Calculate the winding number, denoted as W(γ, 0), which represents the number of times γ winds counterclockwise around the origin. The winding number is an integer.
Linking Winding Number and Roots: According to the Argument Principle and the Fundamental Theorem of Algebra, the winding number W(γ, 0) is equal to the total number of roots of P(z) enclosed by γ, counting multiplicities.
Assumption Contradiction: We assumed that P(z) has no complex roots, which means there are no roots enclosed by any closed path γ in the complex plane.
Contradiction: However, if there are no roots enclosed by any path γ, then the winding number W(γ, 0) must be zero for all such paths. This leads to a contradiction since the winding number is a topological invariant and should not change for all paths that encircle the origin.
Conclusion: The contradiction arises from our initial assumption that there exists a non-constant polynomial equation with complex coefficients that has no complex roots. Therefore, this assumption must be false.

This proof uses the concept of winding numbers and topological properties of paths in the complex plane to establish the existence of complex roots for non-constant polynomial equations. It avoids the explicit use of fixed points and provides an alternative topological perspective on the Fundamental Theorem of Algebra.

Here is a live demonstration of the topological proofs.

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