9.4: Fundamental Theorem of Algebra
Algebra over the complex numbers is in many ways easier than over the real numbers. The reason is that a polynomial of degree \(N\) in \(\mathbb{C}[z]\) has exactly \(N\) zeroes, counting multiplicity. This is called the Fundamental Theorem of Algebra. To prove it, we must establish some preliminary results.
Some Analysis.
DEFINITION. We say that a sequence \(\left\langle z_{n}=x_{n}+i y_{n}\right\rangle\) of complex numbers converges to the number \(z=x+i y\) iff \(\left\langle x_{n}\right\rangle\) converges to \(x\) and \(\left\langle y_{n}\right\rangle\) converges to \(y\) . We say the sequence is Cauchy iff both \(\left\langle x_{n}\right\rangle\) and \(\left\langle y_{n}\right\rangle\) are Cauchy.
REMARK. This is the same as saying that \(\left\langle z_{n}\right\rangle\) converges to \(z\) iff \(\left|z-z_{n}\right|\) tends to zero, and that \(\left\langle z_{n}\right\rangle\) is Cauchy iff \[(\forall \varepsilon>0)(\exists N)(\forall m, n>N)\left|z_{m}-z_{n}\right|<\varepsilon .\] DEFINITION. Let \(G \subseteq \mathbb{C}\) . We say a function \(f: G \rightarrow \mathbb{C}\) is continuous on \(G\) if, whenever \(\left\langle z_{n}\right\rangle\) is a sequence in \(G\) that converges to some value \(z_{\infty}\) in \(G\) , then \(\left\langle f\left(z_{n}\right)\right\rangle\) converges to \(f\left(z_{\infty}\right)\) .
PROPOSITION 9.34. Polynomials are continuous functions on \(\mathbb{C}\) .
Proof. Repeat the proof of Proposition \(5.23\) with complex numbers instead of real numbers.
DEFINITION. A closed rectangle is a set of the form \(\{z \in \mathbb{C} \mid a \leq\) \(\Re(z) \leq b, c \leq \Im(z) \leq d\}\) for some real numbers \(a \leq b\) and \(c \leq d\) . We would like a version of the Extreme Value Theorem, but it is not clear how the minimum and maximum values of a complex valued function should be defined. However, our definition of continuity makes sense even if the range of \(f\) is contained in \(\mathbb{R}\) , and every complex valued continuous function \(g\) has three naturally associated real-valued continuous functions, viz. \(\Re(g), \Im(g)\) and \(|g|\) .
THEOREM 9.35. Let \(R\) be a closed rectangle in \(\mathbb{C}\) , and \(f: R \rightarrow \mathbb{R} a\) continuous function. Then \(f\) attains its maximum and its minimum.
Proof. Let \(R=\{z \in \mathbb{C} \mid a \leq \Re(z) \leq b, c \leq \Im(z) \leq d\}\) . Let \(\left\langle z_{n}=x_{n}+i y_{n}\right\rangle\) be a sequence of points such that \(f\left(z_{n}\right)\) tends to either the least upper bound of the range of \(f\) , if this exists, or let \(f\left(z_{n}\right)>n\) for all \(n\) , if the range is not bounded above. By the BolzanoWeierstrass Theorem 8.6, there is some subsequence for which the real parts converge to some number \(x_{\infty}\) in \([a, b]\) . By Bolzano-Weierstrass again, some subsequence of this subsequence has the property that the imaginary parts also converge, to some point \(y_{\infty}\) in \([c, d]\) . So, replacing the original sequence by this subsequence of the subsequence, we can assume that \(z_{n}\) converges to the point \(z_{\infty}=x_{\infty}+i y_{\infty} \in R\) . By continuity, \(f\left(z_{\infty}\right)=\lim _{n \rightarrow \infty} f\left(z_{n}\right)\) . If the original sequence were unbounded then \(f\left(z_{n}\right)>n\) in the subsequence. This is impossible since the sequence \(\left\langle f\left(z_{n}\right)\right\rangle\) converges to \(f\left(z_{\infty}\right)\) . Therefore the subsequence is bounded and \(f\left(z_{\infty}\right)\) must be the least upper bound of the range of \(f\) . Therefore \(f\left(z_{\infty}\right)\) is the maximum of \(f\) over \(R\) .
A similar argument shows that the minimum is also attained.
REMARK. The previous theorem can be improved to show that a continuous real-valued function on a closed bounded set in \(\mathbb{C}\) attains its extrema. A set \(F\) is closed if whenever a sequence of points \(\left\langle z_{n}\right\rangle\) converges to some complex number \(z_{\infty}\) , then \(z_{\infty}\) is in \(F\) . A set is bounded if it is contained in some rectangle.
We need one more geometric fact. LEMMA 9.36. Triangle inequality Let \(z_{1}, z_{2}\) be complex numbers. Then \[\left|z_{1}+z_{2}\right| \leq\left|z_{1}\right|+\left|z_{2}\right|\]
Proof. Write \(z_{1}=r_{1} \operatorname{Cis}\left(\theta_{1}\right)\) and \(z_{2}=r_{2} \operatorname{Cis}\left(\theta_{2}\right)\) . Then \[\begin{aligned} \left|r_{1} \operatorname{Cis}\left(\theta_{1}\right)+r_{2} \operatorname{Cis}\left(\theta_{2}\right)\right| \\ &=\left[\left(r_{1} \cos \theta_{1}+r_{2} \cos \theta_{2}\right)^{2}+\left(r_{1} \sin \theta_{1}+r_{2} \operatorname{si}\right.\right.\\ &=\left[r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2}\left(\cos \theta_{1} \cos \theta_{2}+\sin \theta_{1} \sin \theta_{2}\right)\right] \\ &=\left[r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)\right]^{1 / 2} \\ & \leq\left[r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2}\right]^{1 / 2} \\ &=r_{1}+r_{2} . \end{aligned}\] COROLLARY 9.38. Let \(z_{1}, \ldots, z_{n} \in \mathbb{C}\) . Then \[\left|z_{1}+\cdots+z_{n}\right| \leq\left|z_{1}\right|+\cdots+\left|z_{n}\right| .\]
The Proof of the Fundamental Theorem of Algebra.
First we observe that finding roots and finding factors are closely related.
LEMMA 9.39. Let \(p\) be a polynomial of degree \(N \geq 1\) in \(\mathbb{C}[z]\) . A complex number, \(c\) , is a root of \(p\) iff \[p(z)=(z-c) q(z),\] where \(q\) is a polynomial of degree \(N-1\) .
Proof. Repeat the proof of Lemma \(4.13\) with real numbers replaced by complex numbers.
Now we prove D’Alembert’s lemma, which states that the modulus of a polynomial cannot have a local minimum except at a root.
LEMMA 9.40. D’Alembert’s Lemma Let \(p \in \mathbb{C}[z]\) and \(\alpha \in \mathbb{C}\) . If \(p(\alpha) \neq 0\) , then \[(\forall \varepsilon>0)(\exists \zeta)[|\zeta-\alpha|<\varepsilon] \wedge[|p(\zeta)|<|p(\alpha)|]\] Proof. Fix \(\alpha\) , not a root of \(p\) . Write \(p\) as \[p(z)=\sum_{k=0}^{N} a_{k}(z-\alpha)^{k},\] where neither \(a_{0}\) nor \(a_{N}\) are 0 . Let \[m=\min \left\{j \in \mathbb{N}^{+} \mid a_{j} \neq 0\right\} .\] So \[p(z)=a_{0}+a_{m}(z-\alpha)^{m}+\cdots+a_{N}(z-\alpha)^{N} .\] Let \(a_{0}=r_{0} \operatorname{Cis}\left(\theta_{0}\right)\) and \(a_{m}=r_{m} \operatorname{Cis}\left(\theta_{m}\right)\) . We will choose \(\zeta\) of the form \[\zeta=\alpha+\rho \operatorname{Cis}(\phi)\] in such a way as to get some cancellation in the first two terms of (9.42). So, let \[\phi=\frac{\theta_{0}+\pi-\theta_{m}}{m} .\] Then \[a_{0}+a_{m}(\zeta-\alpha)^{m}=r_{0} \operatorname{Cis}\left(\theta_{0}\right)-r_{m} \rho^{m} \operatorname{Cis}\left(\theta_{0}\right) .\] It remains to show that, for \(\rho\) small enough, we can ignore all the higher order terms. Note that if \(\rho<1\) , we have \[\begin{aligned} \left|a_{m+1}(\zeta-\alpha)^{m+1}+\cdots+a_{N}(\zeta-\alpha)^{N}\right| \\ & \leq\left|a_{m+1}(\zeta-\alpha)^{m+1}\right|+\cdots+\left|a_{N}(\zeta-\alpha)^{N}\right| \\ &=\left|a_{m+1}\right| \rho^{m+1}+\cdots+\left|a_{N}\right| \rho^{N} \\ & \leq \rho^{m+1}\left[\left|a_{m+1}\right|+\cdots+\left|a_{N}\right|\right] \\ &=: C \rho^{m+1} . \end{aligned}\] Choose \(\rho\) so that \(r_{m} \rho^{m}<r_{0}\) . Then \[p(\zeta)=\left(r_{0}-r_{m} \rho^{m}\right) \operatorname{Cis}\left(\theta_{0}\right)+a_{m+1}(\zeta-\alpha)^{m+1}+\cdots+a_{N}(\zeta-\alpha)^{N},\] \(\mathrm{SO}\) \[|p(\zeta)| \leq r_{0}-r_{m} \rho^{m}+C \rho^{m+1} .\] If \(\rho<r_{m} / C\) , the right-hand side of (9.43) is smaller than \(r_{0}\) .
So we conclude that by taking \[\rho=\frac{1}{2} \min \left(1, \frac{r_{m}}{C},\left[\frac{r_{0}}{r_{m}}\right]^{1 / m}, \varepsilon\right)\] then \[\zeta=\rho \operatorname{Cis}\left(\frac{\theta_{0}+\pi-\theta_{m}}{m}\right)\] satisfies the conclusion of the lemma.
THEOREM 9.44. Fundamental Theorem of Algebra Let \(p \in \mathbb{C}[z]\) be a polynomial of degree \(N \geq 1\) . Then \(p\) can be factored as \[p(z)=c\left(z-\alpha_{1}\right) \ldots\left(z-\alpha_{N}\right)\] for complex numbers \(c, \alpha_{1}, \ldots, \alpha_{N}\) . Moreover the factoring is unique up to order. Proof. (i) Show that \(p\) has at least one root.
Let \(p(z)=\sum_{k=0}^{N} a_{k} z^{k}\) , with \(a_{N} \neq 0\) . Let \(S\) be the closed square \(\{z \in \mathbb{C} \mid-L \leq \Re(z) \leq L,-L \leq \Im(z) \leq L\}\) , where \(L\) is some (large) number to be chosen later.
If \(|z|=R\) then \[\left|\sum_{k=0}^{N-1} a_{k} z^{k}\right| \leq \sum_{k=0}^{N-1}\left|a_{k}\right| R^{k} .\] Choose \(L_{0}\) so that if \(R \geq L_{0}\) , then \[\sum_{k=0}^{N-1}\left|a_{k}\right| R^{k} \leq \frac{1}{2}\left|a_{N}\right| R^{N} .\] Then if \(L \geq L_{0}\) and \(z\) is outside \(S\) , we have \[\begin{aligned} \left|a_{N} z^{N}\right| &=\left|p(z)-\sum_{k=0}^{N-1} a_{k} z^{k}\right| \\ & \leq|p(z)|+\left|\sum_{k=0}^{N-1} a_{k} z^{k}\right| \\ & \leq|p(z)|+\frac{1}{2}\left|a_{N}\right| L^{N}, \end{aligned}\] where the first inequality is the triangle inequality, and the second because \(|z|>L\) . Choose \(L_{1}\) such that \[\frac{1}{2}\left|a_{N}\right| L_{1}^{N}>\left|a_{0}\right| .\] Let \(L=\max \left(L_{0}, L_{1}\right)\) , and let \(S\) be the corresponding closed square. The function \(|p|\) is continuous on \(S\) , so it attains its minimum at some point, \(\alpha_{1}\) say, by Theorem 9.35. On the boundary of \(S\) , we know \[|p(z)| \geq \frac{1}{2}\left|a_{N}\right| L^{N}>\left|a_{0}\right|=|p(0)| .\] Therefore \(\alpha_{1}\) must be in the interior of \(S\) . By D’Alembert’s lemma, we must have \(p\left(\alpha_{1}\right)=0\) , or else there would be a nearby point \(\zeta\) , also in \(S\) , where \(|p(\zeta)|\) was smaller than \(\left|p\left(\alpha_{1}\right)\right|\) . So \(\alpha_{1}\) is a root of \(p\) .
(ii) Now we apply Lemma \(9.39\) to conclude that we can factor \(p\) as \[p(z)=\left(z-\alpha_{1}\right) q(z)\] where \(q\) is a polynomial of degree \(N-1\) . By a straightforward induction argument, we can factor \(p\) into \(N\) linear factors.
(iii) Uniqueness is obvious. The number \(c\) is the coefficient \(a_{N}\) . The numbers \(a_{k}\) are precisely the points at which the function \(p\) vanishes, as it follows from Proposition \(9.19\) that the product of finitely many complex numbers can be 0 if and only if one of the numbers is itself 0 .