17. Zeros of polynomials in a complex variable
We conclude our discussion of functions of a complex variable with one of our favorite functions, polynomials. Motivated by the definition of real polynomials of a real variable, we start by defining complex polynomials of a complex variable and their zeros. We will then use these definitions to explore interesting theorems about zeros of real polynomials of a complex variable.
Complex polynomials of a complex variable and their zeros
A complex polynomial in a complex variable $z$ is an expression of the form:
$$p(z)=c_0+c_1z+c_2z^2+.....+c_{n-1}z^{n-1}+c_nz^n$$
where the coefficients $c_0,c_1,c_2,....,c_n$ are complex numbers i.e. $c_i \in \mathbb{C}$ and $n \in \mathbb{N}$.
The degree of $p(z)$ is $n$ (if $c_n \neq 0$).
A complex number $\alpha$ is called a zero of $p(z)$ if $p(\alpha)=0$.
Remark: In some texts, you will see the terms roots and zeros used interchangeably. Strictly speaking, the term roots should be used for equations, and the term zeros should be used for functions.
Consider the real polynomial $p$ in a complex variable $z$: $$p(z)=z^2+2z+2$$
Suppose we want to find the zeros of $p(z)$. We would proceed as follows:
$$\begin{aligned}&z^2+2z+2=0\\ &z^2+2z+1+1=0\\ &(z+1)^2+1=0\\ &(z+1)^2=i^2\end{aligned}$$It follows that $z+1=i$ or $z+1=-i$ solves the equation; the zeros of $p(z)$ are thus $-1+i$ and $-1-i$ . We observe that the two roots are a complex conjugate pair. Is this a coincidence?
Non-real zeros of a real polynomial
The non-real zeros of a polynomial with real coefficients i.e.
$$p(z)=a_0+a_1z+a_2z^2+.....+a_{n-1}z^{n-1}+a_nz^n$$
(where $a_n \neq 0$ and the coefficients $a_i \in \mathbb{R}$)
occur in complex conjugate pairs. That is, if $p(\alpha)=0$ then $p(\overline{\alpha})=0$.
Remark: A real polynomial is a polynomial with real coefficients. Study the proof of Theorem 17.1 below and answer the questions that follow:
Proof of Theorem 17.1
Let $\alpha$ be a non-real zero of a polynomial of degree $n$ with real coefficients i.e.
$$p(\alpha)=a_0+a_1\alpha +a_2\alpha^2+....+ a_{n-1}\alpha^{n-1}+a_n\alpha^n=0$$
We shall now show that the conjugate of $\alpha$ is also a zero of $p(z)$ i.e. $p(\overline{\alpha})=0$.
$$\begin{aligned}p(\overline{\alpha})&=a_0+a_1\overline{\alpha}+a_2(\overline{\alpha})^2+.....+a_{n-1}(\overline{\alpha})^{n-1}+a_n(\overline{\alpha})^n\\ &=a_0+a_1\overline{\alpha}+a_2\overline{\alpha^2}+.....+a_{n-1}\overline{\alpha^{n-1}}+a_n\overline{\alpha^n}\\ &=\overline{a_0}+\overline{a_1} \;\overline{\alpha}+\overline{a_2} \;\overline{\alpha^2}+.....+\overline{a_{n-1}}\;\overline{\alpha^{n-1}}+\overline{a_n}\;\overline{\alpha^n}\\ &=\overline{a_0}+\overline{a_1\alpha}+\overline{a_2\alpha^2}+.....+\overline{a_{n-1}\alpha^{n-1}}+\overline{a_n\alpha^n}\\ &=\overline{a_0+a_1\alpha+a_2\alpha^2+......+a_{n-1}\alpha^{n-1}+a_n\alpha^n}\\ &=\overline{p(\alpha)}\\ &=\overline{0}\\ &=0\end{aligned}$$$\Box$
Questions (Please get your answers checked by your tutor/lecturer)
- Where in the proof did we use the fact that the coefficients $a_0,a_1, a_2 ...a_{n-1},a_n \in \mathbb{R}$?
- What property of the complex conjugate makes the expressions at (1) and (2) equal?
- Explain why the expressions in (2) and (3) are equal.
- What property of the complex conjugate makes expressions at (3) and (4) equal?
- What property of the complex conjugate makes the expressions at (4) and (5) equal?
- Explain why expressions in (5) and (6) are equal.
- Where in the proof do we use the fact that $\alpha$ is a zero of $p(z)$?
Generally if you are asked to write out this proof you will also have to be clear about why you were allowed to take each step using the properties of complex numbers that we've defined throughout this chapter.
For an example, suppose we are asked to find a polynomial of the form:
$$p(z)=z^2+bz+c \qquad b,c \in \mathbb{R}$$
and are given that $1+i$ is one of the zeros of $p(z)$, how can we use Theorem 17.1 to help us solve the problem?
Since $p(z)$ has real coefficients, we know that the conjugate of $1+i$ is also a zero of $p(z)$, so we have $p(1+i)=p(1-i)=0$ which means $z-(1+i)$ and $z-(1-i)$ are both factors of $p(z)$.
$$\begin{aligned}p(z)&=(z-(1+i))(z-(1-i))\\ &=((z-1)-i)((z-1)+i)\\ &=(z-1)^2-i^2\\ &=z^2-2z+1+1\\ &=z^2-2z+2 \quad \Box\end{aligned}$$Before we upgraded our number system from one dimension to two dimensions, we used to have polynomials that had no (real) zeros. Now that we allow non-real zeros, can a polynomial have no zeros at all? The Fundamental Theorem of Algebra guarantees us that a non-constant polynomial will always have at least one zero.
The Fundamental Theorem of Algebra
Every non-constant polynomial with complex coefficients has at least one complex zero.
Remark: We will not show the proof of the FTA in this course, you'll probably see the proof in MAM3000W. It is believed that Gauss provided the first proof of this theorem in his doctoral dissertation in 1799. Although he only proved it for polynomials with real coefficients, it was not difficult to extend the theorem for polynomials with complex coefficients. Several proofs of the FTA have since been devised in several areas of mathematics including topology and complex analysis (read some proofs here). Some mathematicians actually think that the term "The Fundamental Theorem of Algebra" is inaccurate because the proposition is actually part of complex analysis, calling it "The Fundamental Theorem of Complex Numbers".
The FTA actually tells us something about the number of zeros of a polynomial of degree $n$.
Corollary of the fundamental theorem of algebra
Any polynomial of degree $n \geq 1$ has $n$ zeros counted with multiplicity.
Remark: In some texts, you might see the FTA stated exactly the same way Corollary 17.1 is stated. Counting roots with multiplicity means that the polynomial $p(z)=z^2+2z+1=(z+1)(z+1)$ has two repeated zeros namely $-1$ and $-1$, so when a zero is repeated $n$ times, we count it $n$ times. Corollary 17.1 can also be stated as "Any non-constant polynomial of degree $n$ has at most $n$ distinct zeros".
Proof of Corollary 17.1:
Let $P_n(z)$ be a polynomial of degree $n \geq 1$. By the Fundamental Theorem of Algebra (Theorem 17.2), $P_n(z)$ must have at least one zero, say $r_1$.
Since $r_1$ is a zero of $P_n(z)$, by the factor theorem, $(z-r_1)$ is a factor of $P_n(z)$. Implying that:
$P_n(z)=(z-r_1)Q(z)$ where $Q(z)$ is a polynomial of degree $n-1$.
If $n-1>0$, then $Q(z)$ has at least one zero (by the FTA), say $r_2$.
Continuing in this fashion one may extract $n$ factors of $P(z)$
$P_n(z)=c(z-r_1)(z-r_2)(z-r_3) \cdots (z-r_{n-1})(z-r_n)$ where $c$ is a constant to be determined.
Thus $P_n(z)$ has $n$ zeros counted with multiplicity.
Remark: It is important to bear in mind that the zeros $r_i \in \mathbb{C}$ are not necessarily distinct.
In the earlier sections of the semester when we used partial fraction decomposition to evaluate integrals, we assumed that we could factor every real polynomial into linear and/or irreducible quadratic factors. We are now ready to prove this!
Every non-constant real polynomial can be written as the product of real linear polynomial(s) and/or irreducible real quadratic polynomial(s).
Proof of Theorem 17.3
Consider any zero of a real polynomial $p(z)$ call it $R$.
If $R$ is real then $z-R$ is a real linear factor.
If $R$ is non-real then $\overline{R}$ is also a zero of $p(z)$ (Theorem 17.1) , then $(z-R)$ and $(z-\overline{R})$ are factors of $p(z)$.
So $(z-R)(z-\overline{R})$ is a factor.
$$\begin{aligned}\Big(z-R\Big)\Big(z-\overline{R}\Big)&=z^2-\overline{R}z-Rz+R\overline{R}\\ &=z^2-(\overline{R}+R)z+R\overline{R}\\ &=z^2-2Re(R)z+|R|^2 && \textbf{Properties \href{../sec8-conjugate-and-modulus/#prop-8-1}{8.1}(e) and \href{../sec8-conjugate-and-modulus/#prop-8-1}{8.1}(d)}\end{aligned}$$$z^2-2Re(R)z+|R|^2$ is a real quadratic factor of $p(z)$
The discriminant of this quadratic is:
$$\begin{aligned}\Delta &=(-2Re(R))^2-4(1)(|R|^2)\\ &=4(Re(R))^2-4|R|^2\\ &=4(Re(R))^2-4((Re(R))^2+(Im(R))^2) && \textbf{Definition \href{../sec7-identities-inverses-and-division/#defn-7-2}{7.2} (a)}\\ &=-4(Im(R))^2\\ &< 0 && \textbf{R is not real, so $Im(R) \neq 0$}\end{aligned}$$So $z^2-2Re(R)z+|R|^2$ is an irreducible real quadratic factor.
This means that a polynomial of degree $n \geq 1$ with real coefficients can be factorized into real linear factors and/or real irreducible quadratic factors.
- Given that $p(2-i)=0$, write $p(z)=2z^3-9z^2+14z-5$ as a product of a linear factor and a real irreducible quadratic factor. Find the real zero of $p(z)$.
- Consider the function $g(z)=z^4-2z^3+2z^2-10z+25$. Given that $2+i$ is one of the zeros of $g(z)$, factorize $g$ completely.
Show solutions
- $p(z)$ is a polynomial with real coefficients, therefore all non-real roots come in conjugate pairs. Thus if $p(2-i)=0$, we must have that $(z-(2-i))$ and $(z-(2+i))$ are factors. Given this, then: $$(z-(2-i))(z-(2+i))=z^2-4z+5$$ is a factor. We then know that this must divide $p(z)$. Doing polyonomial long division we find: $$\frac{2z^3-9z^2+14z-5}{5-4z+z^2}=2z-1$$ Therefore $(2z-1)=2(z-\frac{1}{2})$ is a factor, so the real root of this cubic is $\frac{1}{2}$. We can write $$ p(z)= (z^2-4z+5)(2z-1)$$
- As in the previous example $(z-(2+i))(z-(2-i))=z^2-4z+5$ must be a factor of $g$. Again, dividing $g(z)$ by this factor gives: $$\frac{z^4-2z^3+2z^2-10z+25}{z^2-4z+5}=z^2+2z+5$$ This quadratic is factored by completing the square to give two more roots: $z=-1-2i$ and $z=-1+2i$, thus the fully factored polynomial is $$g(z)=(z-(2+i))(z-(2-i))(z-(-1-2i))(z-(-1+2i))$$
- Find all four zeros of $p(z)=z^4+1$ and use them to deduce the factorization $z^4+1=(z^2-\sqrt{2}z+1)(z^2+\sqrt{2}z+1).$
- The polynomial $p(z)=z^4-5z^3+2z^2+22z-20$ has a zero at $z=3-i$. Factorize $p(z)$ as a product of real linear and real irreducible quadratic factors.
Let $q(z)=z^4+1$ and $p(z)=z^5-3z^4+z-3$.
- Find all the complex zeros of $q(z)$ and plot them on an Argand diagram.
- Show that there is a polynomial $r(z)$ such that $p(z)=r(z)q(z)$ and hence find all the zeros of $p(z)$.\
- Factorize $p(z)$ completely into linear and real irreducible quadratic factors.
- Two of the solutions of a cubic equation (with real coefficients) in the complex variable $z$ , are $1+2i$ and $2$. If the constant term is equal to $-30$, find the coefficient of $z^3$.
- Given that $3+i$ is a solution of the equation $$z^4-6z^3+6z^2+24z-40=0$$ find the other solutions of the equation.
- Given that $i$ is a zero of the polynomial $p(z)=z^4+4z^3+14z^2+4z+13$ , find all complex zeros of $p(z)$.
- The complex number $2+i$ is a zero of the polynomial $p(z)=3z^3-14z^2+23z-10$. Use this information to factorize the real-valued polynomial $p(x)=3x^3-14x^2+23x-10$ into real linear and irreducible quadratic factors.