## Section4.2Multiplicative 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 $$|z| = \sqrt{a^2 + b^2}\text{.}$$

###### 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\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 4.17.

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 4.18. 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 $$0 \leq \theta \lt2 \pi\text{.}$$

###### 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}\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.

###### Example4.21.

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

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 ]\text{.} \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\text{.} \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 circle group,

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

The following proposition is a direct result of Proposition 4.20.

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.

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 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 primitive $$n$$th root of unity.

###### 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\text{.} \end{align*}