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Section4.2Multiplicative Group of Complex Numbers

¶

The are defined as

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

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

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

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

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

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

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

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

Example4.16

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

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

and

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

Also,

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

Figure4.17Rectangular coordinates of a complex number

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 or representation. The rectangular representations of $z_1 = 2 + 3i\text{,}$ $z_2 = 1 - 2i\text{,}$ and $z_3 = - 3 + 2i$ are depicted in Figure 4.17.

Figure4.18Polar coordinates of a complex number

Nonzero complex numbers can also be represented using . 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 4.18. We can see that

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

Hence,

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

and

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

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 $0 \leq \theta \lt2 \pi\text{.}$

Example4.19

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

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

and

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

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

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

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.

Proposition4.20

Let $z = r \cis \theta$ and $w = s \cis \phi$ be two nonzero complex numbers. Then

$$\begin{equation*} zw = r s \cis( \theta + \phi). \end{equation*}$$

Example4.21

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

Theorem4.22DeMoivre

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

$$\begin{equation*} [r \cis \theta ]^n = r^n \cis( n \theta) \end{equation*}$$

for $n = 1, 2, \ldots\text{.}$

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

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

Example4.23

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:

$$\begin{align*} z^{10} & = (1+i)^{10}\\ & = \left( \sqrt{2} \cis \left( \frac{\pi }{4} \right) \right)^{10}\\ & = ( \sqrt{2}\, )^{10} \cis \left( \frac{5\pi }{2} \right)\\ & = 32 \cis \left( \frac{\pi }{2} \right)\\ & = 32i. \end{align*}$$

SubsectionThe 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 ,

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

The following proposition is a direct result of Proposition 4.20.

Proposition4.24

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

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 .

Theorem4.25

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

$$\begin{equation*} z = \cis\left( \frac{2 k \pi}{n } \right), \end{equation*}$$

where $k = 0, 1, \ldots, n-1\text{.}$ Furthermore, the $n$th roots of unity form a cyclic subgroup of ${\mathbb T}$ of order $n$

Proof

By DeMoivre's Theorem,

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

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 17.9, 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 .

Example4.26

The 8th roots of unity can be represented as eight equally spaced points on the unit circle (Figure 4.27). The primitive 8th roots of unity are

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

Figure4.278th roots of unity