## Section4.3Cyclic Groups of Complex Numbers

The **complex numbers** are defined as

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. \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 **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 \(|z| = \sqrt{a^2 + b^2}\text{.}\)

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*}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 Figure 4.3.2.

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 Figure 4.3.3. We can see that

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 \(0 \leq \theta \lt2 \pi\text{.}\)

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.

###### Proposition4.3.5

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

###### Theorem4.3.7DeMoivre

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 \(n = 1, 2, \ldots\text{.}\)

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*}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*}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**,

The following proposition is a direct result of Proposition Proposition 4.3.5.

###### Proposition4.3.9

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 **\(n\)th roots of unity**.

###### Theorem4.3.10

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 \(n\)

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 \(2 \pi\text{.}\) 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**.

The 8th roots of unity can be represented as eight equally spaced points on the unit circle (Figure Figure 4.3.12). 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*}