Complex polynomials

Complex number

A complex number is a number of the form

z=a+bi,\qquad a,b\in\mathbb R,\qquad i^2=-1.

The expression $a+bi$ is the Cartesian form of $z$ . Its real part is $\operatorname{Re}(z)=a$ and its imaginary part is $\operatorname{Im}(z)=b$ .

[1]:Argand plane
[2]:Complex conjugate
[3]:Complex modulus
[4]:Add, multiply and divide complex numbers
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Complex polynomial

A complex polynomial is a function $p:\mathbb C\to\mathbb C$ of the form

p(z)=a_0+a_1z+a_2z^2+\cdots+a_nz^n, \qquad a_0,\ldots,a_n\in\mathbb C,

for a non-negative integer $n$ .

The coefficients may be real or non-real. The word polynomial restricts the exponents of the variable to non-negative integers, not the kinds of numbers allowed as coefficients.

Degree

For a non-zero polynomial

p(z)=\sum_{k=0}^n a_kz^k,

the degree of $p$ is the largest integer $k$ for which $a_k\ne0$ . The zero polynomial has undefined degree.

Classify and find degrees

The expressions

0,\quad z,\quad z^{999},\quad 1+35z^4,\quad i+\sqrt2\,iz^4+\pi z

are complex polynomials. The non-zero polynomials $z$ , $z^{999}$ , $1+35z^4$ and $i+\sqrt2\,iz^4+\pi z$ have degrees $1$ , $999$ , $4$ and $4$ , respectively. The zero polynomial has undefined degree.

An expression containing a fractional power such as $z^{1/2}$ is not a polynomial in $z$ . By contrast, numbers such as $2+3i$ and $(1-7i)^{17}$ are allowed as coefficients: only the powers of the variable must be non-negative integers.

[routine]

[1]:Complex polynomial
[2]:Degree

Polynomial root

A number $\alpha\in\mathbb C$ is a root of $p$ if

p(\alpha)=0.

Polynomial factor

A polynomial $p_1$ is a factor of $p$ if there is a polynomial $p_2$ such that

p(z)=p_1(z)p_2(z)

for every $z\in\mathbb C$ .

Find and use roots

For $p(z)=z^2+1$ ,

p(i)=i^2+1=0,

so $i$ is a root. Direct substitution also gives $p(-i)=0$ . The factor theorem below explains why these roots yield the factorisation

z^2+1=(z-i)(z+i)

into linear factors. The same polynomial has no real root: for real $x$ , $x^2+1>0$ .

[routine]

[1]:Polynomial root
[2]:Polynomial factor

Root multiplicity

For a non-zero polynomial $p$ , a root $\alpha$ has multiplicity $k\ge1$ if $(z-\alpha)^k$ is a factor of $p$ but $(z-\alpha)^{k+1}$ is not.

Multiplicity two

For $p(z)=(z-1)^2$ , the root $1$ has multiplicity two: $(z-1)^2$ is a factor of $p$ but $(z-1)^3$ is not.

Multiplicity records repetition in a factorisation. It is not the number of different solutions: this polynomial has one distinct root, counted twice when roots are counted with multiplicity.

[routine]

[1]:Root multiplicity

Roots and factors

Remainder theorem

When $p(z)$ is divided by $z-\alpha$ , the remainder is the complex number $p(\alpha)$ .

Polynomial division gives

p(z)=(z-\alpha)q(z)+r,

with a constant remainder $r$ . At $z=\alpha$ , the first term vanishes, leaving $p(\alpha)=r$ .

[1]:Remainder theorem

Factor theorem

A number $\alpha\in\mathbb C$ is a root of $p$ if and only if $z-\alpha$ is a factor of $p(z)$ .

[1]:Polynomial root
[2]:Polynomial factor
[3]:Remainder theorem
[4]:Proof
[5]:Uniqueness of polynomial coefficients
[6]:Factorisation over the complex numbers
[7]:A complex conjugate pair of roots
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If $p(\alpha)=0$ , polynomial division gives

p(z)=(z-\alpha)q(z)+p(\alpha)=(z-\alpha)q(z),

so $z-\alpha$ is a factor. Conversely, if $p(z)=(z-\alpha)q(z)$ , then $p(\alpha)=0$ .

[1]:Factor theorem

Uniqueness of polynomial coefficients

Two complex polynomials agree for every $z\in\mathbb C$ if and only if their corresponding coefficients are equal. In particular, a complex polynomial is zero for every $z$ if and only if all its coefficients are zero.

This is why identities between polynomial functions can be checked coefficient by coefficient.

To see why, subtract the two polynomials. If the result were a non-zero polynomial of degree $d$ , it could have at most $d$ distinct roots: each root supplies a linear factor by the factor theorem. After dividing out $z-\alpha$ , any other root $\beta\ne\alpha$ remains a root of the quotient, so we can repeat this step at most $d$ times. It therefore cannot vanish at every complex number.

[1]:Complex polynomial
[2]:Factor theorem

Fundamental theorem of algebra

Every complex polynomial of degree $n\ge1$ has at least one complex root.

The theorem is stated without proof. It guarantees that a root exists, not a formula for finding it.

Applying the factor theorem to that root and then repeating with the quotient, gives the full factorisation below. This is what is special about $\mathbb C$ : over $\mathbb R$ , a polynomial such as $z^2+1$ cannot be factored into real linear factors.

[1]:Complex polynomial
[2]:Factorisation over the complex numbers
[3]:Roots of a real polynomial
[4]:Seen questions
[5]:Seen questions
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Factorisation over $\mathbb C$

Factorisation over the complex numbers

Every complex polynomial of degree $n\ge1$ can be written

p(z)=a(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n),

where $a$ is the coefficient of $z^n$ and the $\alpha_j\in\mathbb C$ are roots, counted with multiplicity.

Proceed by induction on the degree. A degree-one polynomial already has the required form. Suppose the result holds for degree $m$ . If $p$ has degree $m+1$ , the fundamental theorem of algebra supplies a root $\alpha_1$ . The factor theorem gives

p(z)=(z-\alpha_1)q(z)

for a polynomial $q$ of degree $m$ . Apply the induction hypothesis to $q$ and multiply the factors.

[1]:Factorisation over the complex numbers

[COROLLARY]

Count roots with multiplicity

A complex polynomial of degree $n\ge1$ has exactly $n$ complex roots counted with multiplicity.

The factorisation theorem gives $n$ linear factors. Conversely, if $p(\beta)=0$ , then

0=a(\beta-\alpha_1)\cdots(\beta-\alpha_n),

so $\beta$ equals one of the $\alpha_j$ . Thus the list has no missing roots, and repeated factors record multiplicity.

“Exactly $n$ roots” does not mean $n$ distinct values: $(z-1)^2$ has two roots counted with multiplicity, but only one distinct root.

[1]:Factorisation over the complex numbers

Factorise a quadratic

For $p(z)=z^2+z+1$ , the quadratic formula gives

z=\frac{-1\pm\sqrt{-3}}2 = -\frac12\pm\frac{\sqrt3}{2}i.

Therefore

z^2+z+1 =\left(z+\frac12-\frac{\sqrt3}{2}i\right) \left(z+\frac12+\frac{\sqrt3}{2}i\right).

[routine]

[1]:Factorisation over the complex numbers

Factorise $z^8-1$

The roots satisfy $z^8=1$ , so the eighth roots of unity give

\alpha_k=e^{2\pi ik/8},\qquad k=0,1,\ldots,7.

Hence

z^8-1=\prod_{k=0}^{7}\left(z-e^{2\pi ik/8}\right)

over $\mathbb C$ .

[exam]

Returning to real factors

A real polynomial is one whose coefficients all lie in $\mathbb R$ , even when the variable takes complex values. Its non-real roots occur in conjugate pairs, so its complex factorisation can be regrouped into real factors.

[PROPOSITION]

Conjugate roots for real coefficients

If $p$ has real coefficients and $\alpha$ is a root of multiplicity $k$ , then $\overline\alpha$ is also a root of multiplicity $k$ .

The condition on the coefficients is essential: conjugation must leave every coefficient unchanged before it can pass through the polynomial.

Write $p(z)=\sum_{j=0}^n a_jz^j$ with $a_j\in\mathbb R$ . If $p(\alpha)=0$ , then

p(\overline\alpha) =\sum_{j=0}^n a_j\overline\alpha^{\,j} =\sum_{j=0}^n\overline{a_j\alpha^j} =\overline{p(\alpha)} =0.

For multiplicity, write $p(z)=(z-\alpha)^kq(z)$ with $q(\alpha)\ne0$ . Since $p$ has real coefficients, conjugating this factorisation gives $p(z)=(z-\overline\alpha)^k\widetilde q(z)$ , where $\widetilde q(z)=\overline{q(\overline z)}$ and $\widetilde q(\overline\alpha)=\overline{q(\alpha)}\ne0$ . Thus the conjugate root has exactly the same multiplicity.

[1]:Conjugate roots for real coefficients

[PROPOSITION]

Real linear and quadratic factors

Every non-zero polynomial with real coefficients can be written as a non-zero real constant times a product of real linear factors and real quadratic factors.

Every non-real root occurs with its conjugate. Pairing the corresponding factors gives

(z-\alpha)(z-\overline\alpha) =z^2-2\operatorname{Re}(\alpha)z+|\alpha|^2,

whose coefficients are real. Real roots already provide real linear factors. The quadratic factors obtained from non-real roots have negative discriminant, so they do not split further over $\mathbb R$ .