## Section6.5Factoring over $\C$ and $\R$

We will now consider how to factor over the larger fields $\R$ and $\C\text{.}$ 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.

### Subsection6.5.1Multiplicative Group of Complex Numbers

The complex numbers are defined as

\begin{equation*} {\mathbb C} = \{ a + bi : a, b \in {\mathbb R} \}\text{,} \end{equation*}

where $i^2 = -1\text{.}$ If $z = a + bi\text{,}$ then $a$ is the real part of $z$ and $b$ is the imaginary part of $z\text{.}$

To add two complex numbers $z=a+bi$ and $w= c+di\text{,}$ we just add the corresponding real and imaginary parts:

\begin{equation*} z + w=(a + bi ) + (c + di) = (a + c) + (b + d)i\text{.} \end{equation*}

Remembering that $i^2 = -1\text{,}$ we multiply complex numbers just like polynomials. The product of $z$ and $w$ is

\begin{equation*} (a + bi )(c + di) = ac + bdi^2 + adi + bci = (ac -bd) +(ad + bc)i\text{.} \end{equation*}

Every nonzero complex number $z = a +bi$ has a multiplicative inverse; that is, there exists a $z^{-1} \in {\mathbb C}^\ast$ such that $z z^{-1} = z^{-1} z = 1\text{.}$ If $z = a + bi\text{,}$ then

\begin{equation*} z^{-1} = \frac{a-bi}{ a^2 + b^2 }\text{.} \end{equation*}

The complex conjugate of a complex number $z = a + bi$ is defined to be $\overline{z} = a- bi\text{.}$ The absolute value or modulus of $z = a + bi$ is $|z| = \sqrt{a^2 + b^2}\text{.}$

###### Example6.30.

Let $z = 2 + 3i$ and $w = 1-2i\text{.}$ Then

\begin{equation*} z + w = (2 + 3i) + (1 - 2i) = 3 + i \end{equation*}

and

\begin{equation*} z w = (2 + 3i)(1 - 2i ) = 8 - i\text{.} \end{equation*}

Also,

\begin{align*} z^{-1} & = \frac{2}{13} - \frac{3}{13}i\\ |z| & = \sqrt{13}\\ \overline{z} & = 2-3i\text{.} \end{align*}

There are several ways of graphically representing complex numbers. We can represent a complex number $z = a +bi$ as an ordered pair on the $xy$ 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 $z_1 = 2 + 3i\text{,}$ $z_2 = 1 - 2i\text{,}$ and $z_3 = - 3 + 2i$ are depicted in Figure 6.31.

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

\begin{equation*} z = a + bi = r( \cos \theta + i \sin \theta)\text{.} \end{equation*}

Hence,

\begin{equation*} r = |z| = \sqrt{a^2 + b^2} \end{equation*}

and

\begin{align*} a & = r \cos \theta\\ b & = r \sin \theta\text{.} \end{align*}

We sometimes abbreviate $r( \cos \theta + i \sin \theta)$ as $r \cis \theta\text{.}$ To assure that the representation of $z$ is well-defined, we also require that $0^{\circ} \leq \theta \lt 360^{\circ}\text{.}$ If the measurement is in radians, then $0 \leq \theta \lt2 \pi\text{.}$

###### Example6.33.

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

\begin{equation*} a = 2 \cos 60^{\circ} = 1 \end{equation*}

and

\begin{equation*} b = 2 \sin 60^{\circ} = \sqrt{3}\text{.} \end{equation*}

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

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

\begin{equation*} r = \sqrt{a^2 + b^2} = \sqrt{36 } = 6 \end{equation*}

and

\begin{equation*} \theta = \arctan \left( \frac{b}{a} \right) = \arctan( - 1) = 315^{\circ}\text{,} \end{equation*}

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

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.

###### Example6.35.

If $z = 3 \cis( \pi / 3 )$ and $w = 2 \cis(\pi / 6 )\text{,}$ then $zw = 6 \cis( \pi / 2 ) = 6i\text{.}$

###### Proof.

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

\begin{align*} z^{n+1} & = z^n z\\ & = r^n( \cos n \theta + i \sin n \theta ) r( \cos \theta + i\sin \theta )\\ & = r^{n+1} [( \cos n \theta \cos \theta - \sin n \theta \sin \theta ) + i ( \sin n \theta \cos \theta + \cos n \theta \sin \theta)]\\ & = r^{n+1} [ \cos( n \theta + \theta) + i \sin( n \theta + \theta) ]\\ & = r^{n+1} [ \cos( n +1) \theta + i \sin( n+1) \theta ]\text{.} \end{align*}
###### Example6.37.

Suppose that $z= 1+i$ and we wish to compute $z^{10}\text{.}$ 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{align*} z^{10} & = (1+i)^{10}\\ & = \left( \sqrt{2} \cis \left( \frac{\pi }{4} \right) \right)^{10}\\ & = ( \sqrt{2}\, )^{10} \cis \left( \frac{5\pi }{2} \right)\\ & = 32 \cis \left( \frac{\pi }{2} \right)\\ & = 32i\text{.} \end{align*}
##### The Circle Group and the Roots of Unity.

The multiplicative group of the complex numbers, ${\mathbb C}^*\text{,}$ possesses some interesting subgroups. Whereas ${\mathbb Q}^*$ and ${\mathbb R}^*$ have no interesting subgroups of finite order, ${\mathbb C}^*$ has many. We first consider the circle group,

\begin{equation*} {\mathbb T} = \{ z \in {\mathbb C} : |z| = 1 \}\text{.} \end{equation*}

The following proposition is a direct result of Proposition 6.34 .

Although the circle group has infinite order, it has many interesting finite subgroups. Suppose that $H = \{ 1, -1, i, -i \}\text{.}$ Then $H$ is a subgroup of the circle group. Also, $1\text{,}$ $-1\text{,}$ $i\text{,}$ and $-i$ are exactly those complex numbers that satisfy the equation $z^4 = 1\text{.}$ The complex numbers satisfying the equation $z^n=1$ are called the $n$th roots of unity.

###### Proof.

By DeMoivre's Theorem,

\begin{equation*} z^n = \cis \left( n \frac{2 k \pi}{n } \right) = \cis( 2 k \pi ) = 1\text{.} \end{equation*}

The $z$'s are distinct since the numbers $2 k \pi /n$ are all distinct and are greater than or equal to 0 but less than $2 \pi\text{.}$ 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 ${\mathbb T}$ as an exercise.

A generator for the group of the $n$th roots of unity is called a primitive $n$th root of unity.

###### Example6.40.

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{align*} \omega & = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\\ \omega^3 & = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\\ \omega^5 & = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i\\ \omega^7 & = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\text{.} \end{align*}

### Subsection6.5.2Factoring With and Without Complex Numbers

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 $\R\text{.}$ What about even degree polynomials? For example, what about $x^2 + 3\text{?}$

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

Let's see how to factor in the complex numbers. It turns out (although hard to prove) that over $\C\text{,}$ every polynomial factors into linear terms. The Fundamental Theorem of Algebra says: Every non-constant polynomial in $\mathbb C[x]$ has a complex root.

What does this tell us about the irreducible polynomials in $\C[x]\text{?}$ They are exactly the degree 1 polynomials. If $a(x)$ is a polynomial for degree greater than 1 in $\C[x]\text{,}$ then it must have a complex root $c\text{,}$ so $x-c$ is a factor. Applying this repeatedly, we find

\begin{equation*} a(x) = k(x-c_1)(x-c_2)\cdots(x-c_n)\text{.} \end{equation*}

There will be exactly $n$ roots (although note that the roots might not be distinct).

What about simple polynomials like $x^5 - 1\text{.}$ Okay, that is not irreducible, since 1 is a root. Does it have any other roots?

###### Example6.42.

Factor $x^5-1\text{.}$

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 $re^{i\theta}$ and took the $n$th root, we would get $\sqrt[n]{r}e^{i\theta/n}\text{.}$

###### Example6.43.

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

Could we factor the polynomial $x^3+5$ over $\R\text{?}$ We could simply use long division to factor out $x+\sqrt{5}\text{,}$ although that might be messy. Notice though that if we did, then the quotient will be a degree 2 polynomial with real coefficients.

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

###### Example6.44.

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

Will something like this always work? If $a+bi$ is a root of $a(x)\text{,}$ then $a-bi$ is also a root of $a(x)$ (in $\C$). That is, if $r$ is a root, so is its complex conjugate $\overline r$ is as well.

How do we know? Well the function $f(r) = \overline r$ is a ring homomorphism from $\mathbb C\to \mathbb C\text{.}$ Check this. But then if $a(r) = 0\text{,}$ apply $f$ to both sides to get $a(\overline r) = 0\text{.}$

This is great: if $x-r$ is a factor of $a(x)$ in $\mathbb C[x]\text{,}$ then so is $x-\overline r\text{.}$ But notice:

\begin{equation*} (x-r)(x-\overline r) = x^2 - 2ax + (a^2 + b^2) \end{equation*}

is a quadratic polynomial with real coefficients.

This shows that every polynomial in $\mathbb R[x]$ can be factored into polynomials of degree 1 or 2 in $\mathbb R[x]\text{.}$

Thus the irreducible polynomials in $\mathbb R[x]$ are exactly the linear polynomials and the quadratics with negative discriminant (i.e., $b^2 - 4ac \lt 0$).

Here is another example.

###### Example6.45.

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

### Exercises6.5.3Practice Problems

###### 1.

Evaluate each of the following.

1. $\displaystyle (3-2i)+ (5i-6)$

2. $\displaystyle (4-5i)-\overline{(4i -4)}$

3. $\displaystyle (5-4i)(7+2i)$

4. $\displaystyle (9-i) \overline{(9-i)}$

5. $\displaystyle i^{45}$

6. $\displaystyle (1+i)+\overline{(1+i)}$

Hint

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

###### 2.

Convert the following complex numbers to the form $a + bi\text{.}$

1. $\displaystyle 2 e^{\pi / 6}$

2. $\displaystyle 5 e^{9\pi/4}$

3. $\displaystyle 3 e^{\pi}$

4. $\displaystyle \frac{1}{2}e^{7\pi/4}$

Hint

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

###### 3.

Change the following complex numbers to polar representation.

1. $\displaystyle 1-i$

2. $\displaystyle -5$

3. $\displaystyle 2+2i$

4. $\displaystyle \sqrt{3} + i$

5. $\displaystyle -3i$

6. $\displaystyle 2i + 2 \sqrt{3}$

Hint

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

###### 4.

Calculate each of the following expressions.

1. $\displaystyle (1+i)^{-1}$

2. $\displaystyle (1 - i)^{6}$

3. $\displaystyle (\sqrt{3} + i)^{5}$

4. $\displaystyle (-i)^{10}$

5. $\displaystyle ((1-i)/2)^{4}$

6. $\displaystyle (-\sqrt{2} - \sqrt{2}\, i)^{12}$

7. $\displaystyle (-2 + 2i)^{-5}$

Hint

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

###### 5.

Prove that the function $\phi:\bC \to \bC$ given by $\phi(z) = \bar z$ is a ring homomorphism. Here $\bar z = a-bi$ is the complex conjugate of $z = a+bi$

###### 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.

###### 7.

Is $x^7 - 1$ irreducible over $\bC\text{?}$ How many roots should it have? Find all of them. Hint: use the polar form of complex numbers, $re^{i\vartheta}\text{.}$

###### 8.

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

### Exercises6.5.4Collected Homework

###### 1.

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

Hint

Do the factoring in that order.

###### 2.

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