# 9.4: Fundamental Theorem of Algebra

• • Bob Dumas and John E. McCarthy
• University of Washington and Washington University in St. Louis
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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 $\left|z_{1}+z_{2}\right| \leq\left|z_{1}\right|+\left|z_{2}\right|$ FIGURE $$9.18$$. Triangle inequality

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 .

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