AATA Factoring over $\C$ and $\R$

Section 6.5 Factoring over $C$ and $R$

We will now consider how to factor over the larger fields

R

and

C .

It will turn out that even if we only care about factoring polynomials over the real numbers, working in the complex numbers is helpful. Thus we begin by reviewing a bit about the complex numbers and describing some of their group structure.

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Subsection 6.5.1 Multiplicative Group of Complex Numbers

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The complex numbers are defined as

C = {a + b i : a, b \in R},

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where

i^{2} = - 1 .

z = a + b i,

then

a

is the real part of

z

and

b

is the imaginary part of

z .

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To add two complex numbers

z = a + b i

and

w = c + d i,

we just add the corresponding real and imaginary parts:

z + w = (a + b i) + (c + d i) = (a + c) + (b + d) i .

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Remembering that

i^{2} = - 1,

we multiply complex numbers just like polynomials. The product of

z

and

w

(a + b i) (c + d i) = a c + b d i^{2} + a d i + b c i = (a c - b d) + (a d + b c) i .

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Every nonzero complex number

z = a + b i

has a multiplicative inverse; that is, there exists a

z^{- 1} \in C^{*}

such that

z z^{- 1} = z^{- 1} z = 1 .

z = a + b i,

then

z^{- 1} = \frac{a - b i}{a^{2} + b^{2}} .

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The complex conjugate of a complex number

z = a + b i

is defined to be

\overset{―}{z} = a - b i .

The absolute value or modulus of

z = a + b i

| z | = \sqrt{a^{2} + b^{2}} .

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Example 6.30.

Let $z = 2 + 3 i$ and $w = 1 - 2 i .$ Then

z + w = (2 + 3 i) + (1 - 2 i) = 3 + i

and

z w = (2 + 3 i) (1 - 2 i) = 8 - i .

Also,

\begin{aligned} z^{- 1} & = \frac{2}{13} - \frac{3}{13} i \\ | z | & = \sqrt{13} \\ \overset{―}{z} & = 2 - 3 i . \end{aligned}

The complex plane where the horizontal axis is the x-axis or real axis and the verticle axis is the y-axis or imaginary axis. The point z1 = 2 + 3i is in the upper right quadrant, the point z2 = 1- 2i in the lower right quadrant, and z3 = -3 + 2i in the upper right quadrant. — 🔗
Figure 6.31. Rectangular coordinates of a complex number

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There are several ways of graphically representing complex numbers. We can represent a complex number

z = a + b i

as an ordered pair on the

x y

plane where

a

is the

x

(or real) coordinate and

b

is the

y

(or imaginary) coordinate. This is called the rectangular or Cartesian representation. The rectangular representations of

z_{1} = 2 + 3 i,

z_{2} = 1 - 2 i,

and

z_{3} = - 3 + 2 i

are depicted in Figure 6.31.

The complex plane where the horizontal axis is the x-axis or real axis and the verticle axis is the y-axis or imaginary axis. The point a + bi is in the upper right quadrant. The point is also determined by a ray that at an angle of theta counterclockwise from the horizontal axis having a length of r. — 🔗
Figure 6.32. Polar coordinates of a complex number

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Nonzero complex numbers can also be represented using polar coordinates. To specify any nonzero point on the plane, it suffices to give an angle

θ

from the positive

x

axis in the counterclockwise direction and a distance

r

from the origin, as in Figure 6.32. We can see that

z = a + b i = r (\cos θ + i \sin θ) .

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Hence,

r = | z | = \sqrt{a^{2} + b^{2}}

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and

\begin{aligned} a & = r \cos θ \\ b & = r \sin θ . \end{aligned}

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We sometimes abbreviate

r (\cos θ + i \sin θ)

r cis θ .

To assure that the representation of

z

is well-defined, we also require that

0^{\circ} \leq θ < 360^{\circ} .

If the measurement is in radians, then

0 \leq θ < 2 π .

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Example 6.33.

Suppose that $z = 2 cis 60^{\circ} .$ Then

a = 2 \cos 60^{\circ} = 1

and

b = 2 \sin 60^{\circ} = \sqrt{3} .

Hence, the rectangular representation is $z = 1 + \sqrt{3} i .$

Conversely, if we are given a rectangular representation of a complex number, it is often useful to know the number's polar representation. If $z = 3 \sqrt{2} - 3 \sqrt{2} i,$ then

r = \sqrt{a^{2} + b^{2}} = \sqrt{36} = 6

and

θ = \arctan (\frac{b}{a}) = \arctan (- 1) = 315^{\circ},

so $3 \sqrt{2} - 3 \sqrt{2} i = 6 cis 315^{\circ} .$

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The polar representation of a complex number makes it easy to find products and powers of complex numbers. The proof of the following proposition is straightforward and is left as an exercise.

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Proposition 6.34.

Let $z = r cis θ$ and $w = s cis ϕ$ be two nonzero complex numbers. Then

z w = r s cis (θ + ϕ) .

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Example 6.35.

If $z = 3 cis (π / 3)$ and $w = 2 cis (π / 6),$ then $z w = 6 cis (π / 2) = 6 i .$

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Theorem 6.36. DeMoivre.

Let $z = r cis θ$ be a nonzero complex number. Then

[r cis θ]^{n} = r^{n} cis (n θ)

for $n = 1, 2, \dots .$

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Proof.

We will use induction on $n .$ For $n = 1$ the theorem is trivial. Assume that the theorem is true for all $k$ such that $1 \leq k \leq n .$ Then

\begin{aligned} z^{n + 1} & = z^{n} z \\ = r^{n} (\cos n θ + i \sin n θ) r (\cos θ + i \sin θ) \\ = r^{n + 1} [(\cos n θ \cos θ - \sin n θ \sin θ) + i (\sin n θ \cos θ + \cos n θ \sin θ)] \\ = r^{n + 1} [\cos (n θ + θ) + i \sin (n θ + θ)] \\ = r^{n + 1} [\cos (n + 1) θ + i \sin (n + 1) θ] . \end{aligned}

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Example 6.37.

Suppose that $z = 1 + i$ and we wish to compute $z^{10} .$ Rather than computing $(1 + i)^{10}$ directly, it is much easier to switch to polar coordinates and calculate $z^{10}$ using DeMoivre's Theorem:

\begin{aligned} z^{10} & = (1 + i)^{10} \\ = {(\sqrt{2} cis (\frac{π}{4}))}^{10} \\ = (\sqrt{2})^{10} cis (\frac{5 π}{2}) \\ = 32 cis (\frac{π}{2}) \\ = 32 i . \end{aligned}

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The Circle Group and the Roots of Unity.

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The multiplicative group of the complex numbers,

C^{*},

possesses some interesting subgroups. Whereas

Q^{*}

and

R^{*}

have no interesting subgroups of finite order,

C^{*}

has many. We first consider the circle group,

T = {z \in C : | z | = 1} .

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The following proposition is a direct result of Proposition 6.34 .

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Proposition 6.38.

The circle group is a subgroup of $C^{*} .$

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Although the circle group has infinite order, it has many interesting finite subgroups. Suppose that

H = {1, - 1, i, - i} .

Then

H

is a subgroup of the circle group. Also,

1,

- 1,

i,

and

- i

are exactly those complex numbers that satisfy the equation

z^{4} = 1 .

The complex numbers satisfying the equation

z^{n} = 1

are called the

n

th roots of unity.

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Theorem 6.39.

If $z^{n} = 1,$ then the $n$ th roots of unity are

z = cis (\frac{2 k π}{n}),

where $k = 0, 1, \dots, n - 1 .$ Furthermore, the $n$ th roots of unity form a cyclic subgroup of $T$ of order $n$

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Proof.

By DeMoivre's Theorem,

z^{n} = cis (n \frac{2 k π}{n}) = cis (2 k π) = 1 .

The $z$ 's are distinct since the numbers $2 k π / n$ are all distinct and are greater than or equal to 0 but less than $2 π .$ The fact that these are all of the roots of the equation $z^{n} = 1$ follows from from Corollary 6.16, which states that a polynomial of degree $n$ can have at most $n$ roots. We will leave the proof that the $n$ th roots of unity form a cyclic subgroup of $T$ as an exercise.

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A generator for the group of the

n

th roots of unity is called a primitive

n

th root of unity.

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Example 6.40.

The 8th roots of unity can be represented as eight equally spaced points on the unit circle (Figure 6.41). The primitive 8th roots of unity are

\begin{aligned} ω & = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \\ ω^{3} & = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \\ ω^{5} & = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i \\ ω^{7} & = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i . \end{aligned}

The 8 roots of unity are spaced evenly around the unit circle beginning with 1 on the positive horizontal axis and followed by omega, i, the cube of omega, -1, omega to the fifth power, -i, and omega to the seventh power. — Figure 6.41. 8th roots of unity

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Subsection 6.5.2 Factoring With and Without Complex Numbers

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Every odd degree polynomial has a root in

R

(how do we know?), so no odd degree polynomial (of degree at least 3) can be irreducible over

R .

What about even degree polynomials? For example, what about

x^{2} + 3 ?

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Well, that polynomial has roots, but they are complex roots, in particular, non-real complex roots.

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Let's see how to factor in the complex numbers. It turns out (although hard to prove) that over

C,

every polynomial factors into linear terms. The Fundamental Theorem of Algebra says: Every non-constant polynomial in

C [x]

has a complex root.

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What does this tell us about the irreducible polynomials in

C [x] ?

They are exactly the degree 1 polynomials. If

a (x)

is a polynomial for degree greater than 1 in

C [x],

then it must have a complex root

c,

x - c

is a factor. Applying this repeatedly, we find

a (x) = k (x - c_{1}) (x - c_{2}) \dots (x - c_{n}) .

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There will be exactly

n

roots (although note that the roots might not be distinct).

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What about simple polynomials like

x^{5} - 1 .

Okay, that is not irreducible, since 1 is a root. Does it have any other roots?

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Example 6.42.

Factor $x^{5} - 1 .$

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The example above is perhaps a little simplistic. In particular,

r = 1

here, so we didn't need to do anything with that. In general, if we had

r e^{i θ}

and took the

n

th root, we would get

\sqrt[n]{r} e^{i θ / n} .

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Example 6.43.

Factor $x^{3} + 5$ over the complex numbers.

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Could we factor the polynomial

x^{3} + 5

over

R ?

We could simply use long division to factor out

x + \sqrt[3]{5},

although that might be messy. Notice though that if we did, then the quotient will be a degree 2 polynomial with real coefficients.

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There is a better way. We should be able to get the same polynomial by multiplying the two complex factors from the example above.

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Example 6.44.

Factor $x^{3} + 5$ over the real numbers.

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Will something like this always work? If

a + b i

is a root of

a (x),

then

a - b i

is also a root of

a (x)

(in

C

). That is, if

r

is a root, so is its complex conjugate

\overset{―}{r}

is as well.

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How do we know? Well the function

f (r) = \overset{―}{r}

is a ring homomorphism from

C \to C .

Check this. But then if

a (r) = 0,

apply

f

to both sides to get

a (\overset{―}{r}) = 0 .

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This is great: if

x - r

is a factor of

a (x)

C [x],

then so is

x - \overset{―}{r} .

But notice:

(x - r) (x - \overset{―}{r}) = x^{2} - 2 a x + (a^{2} + b^{2})

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is a quadratic polynomial with real coefficients.

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This shows that every polynomial in

R [x]

can be factored into polynomials of degree 1 or 2 in

R [x] .

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Thus the irreducible polynomials in

R [x]

are exactly the linear polynomials and the quadratics with negative discriminant (i.e.,

b^{2} - 4 a c < 0

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Here is another example.

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Example 6.45.

Factor $p (x) = x^{9} - 2 x^{8} - 3 x^{7} - 2 x^{2} + 4 x + 6$ completely over $Q,$ $R$ and $C .$

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Exercises 6.5.3 Practice Problems

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1.

Evaluate each of the following.

$(3 - 2 i) + (5 i - 6)$
$(4 - 5 i) - \overset{―}{(4 i - 4)}$
$(5 - 4 i) (7 + 2 i)$
$(9 - i) \overset{―}{(9 - i)}$
$i^{45}$
$(1 + i) + \overset{―}{(1 + i)}$

Hint.

(a) $-3 + 3i\text{;}$ (c) $43- 18i\text{;}$ (e) $i$

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2.

Convert the following complex numbers to the form $a + b i .$

$2 e^{π / 6}$
$5 e^{9 π / 4}$
$3 e^{π}$
$\frac{1}{2} e^{7 π / 4}$

Hint.

(a) $\sqrt{3} + i\text{;}$ (c) $-3\text{.}$

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3.

Change the following complex numbers to polar representation.

$1 - i$
$- 5$
$2 + 2 i$
$\sqrt{3} + i$
$- 3 i$
$2 i + 2 \sqrt{3}$

Hint.

(a) $\sqrt{2} \cis( 7 \pi /4)\text{;}$ (c) $2 \sqrt{2} \cis( \pi /4)\text{;}$ (e) $3 \cis(3 \pi/2)\text{.}$

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4.

Calculate each of the following expressions.

$(1 + i)^{- 1}$
$(1 - i)^{6}$
$(\sqrt{3} + i)^{5}$
$(- i)^{10}$
$((1 - i) / 2)^{4}$
$(- \sqrt{2} - \sqrt{2} i)^{12}$
$(- 2 + 2 i)^{- 5}$

Hint.

(a) $(1 - i)/2\text{;}$ (c) $16(i - \sqrt{3}\, )\text{;}$ (e) $-1/4\text{.}$

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5.

Prove that the function $ϕ : C \to C$ given by $ϕ (z) = \bar{z}$ is a ring homomorphism. Here $\bar{z} = a - b i$ is the complex conjugate of $z = a + b i$

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6.

Let $z$ be a complex number. Prove that the sum $z + \bar{z}$ and product $z \cdot \bar{z}$ or the number with its conjugate are always real numbers.

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7.

Is $x^{7} - 1$ irreducible over $C ?$ How many roots should it have? Find all of them. Hint: use the polar form of complex numbers, $r e^{i ϑ} .$

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8.

Factor $x^{8} - 5 x^{7} - 14 x^{6} + x^{2} - 5 x - 14$ completely over $Q,$ $C$ and $R$

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Exercises 6.5.4 Collected Homework

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1.

Factor the polynomial $p (x) = x^{7} + 2 x^{6} - 3 x - 6$ completely (into irreducible factors) over $Q,$ then over $C,$ and then over $R .$

Hint.

Do the factoring in that order.

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2.

True or false: $x^{4} + 20 x^{3} + 5 x^{2} + 10 x + 15$ is irreducible in $R [x] .$ Briefly explain.

Section 6.5 Factoring over C and R

Subsection 6.5.1 Multiplicative Group of Complex Numbers

Example 6.30.

Example 6.33.

Proposition 6.34.

Example 6.35.

Theorem 6.36. DeMoivre.

Proof.

Example 6.37.

The Circle Group and the Roots of Unity.

Proposition 6.38.

Theorem 6.39.

Proof.

Example 6.40.

Subsection 6.5.2 Factoring With and Without Complex Numbers

Example 6.42.

Example 6.43.

Example 6.44.

Example 6.45.

Exercises 6.5.3 Practice Problems

1.

2.

3.

4.

5.

6.

7.

8.

Exercises 6.5.4 Collected Homework

1.

2.

Section 6.5 Factoring over $C$ and $R$