# Section4.2Multiplicative Group of Complex Numbers¶ permalink

The complex numbers are defined as \begin{equation*}{\mathbb C} = \{ a + bi : a, b \in {\mathbb R} \},\end{equation*} where $i^2 = -1$. If $z = a + bi$, then $a$ is the real part of $z$ and $b$ is the imaginary part of $z$.

To add two complex numbers $z=a+bi$ and $w= c+di$, 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 $i^2 = -1$, 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 $z z^{-1} = z^{-1} z = 1$. If $z = a + bi$, 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 $\overline{z} = a- bi$. The absolute value or modulus of $z = a + bi$ is $|z| = \sqrt{a^2 + b^2}$.

##### Example4.16

Let $z = 2 + 3i$ and $w = 1-2i$. 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 $z_1 = 2 + 3i$, $z_2 = 1 - 2i$, 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).\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 $r \cis \theta$. To assure that the representation of $z$ is well-defined, we also require that $0^{\circ} \leq \theta \lt 360^{\circ}$. If the measurement is in radians, then $0 \leq \theta \lt2 \pi$.

##### Example4.19

Suppose that $z = 2 \cis 60^{\circ}$. 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 $z = 1+\sqrt{3}\, i$.

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

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 $w = 2 \cis(\pi / 6 )$, then $zw = 6 \cis( \pi / 2 ) = 6i$.

##### Example4.23

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{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 ¶ permalink

The multiplicative group of the complex numbers, ${\mathbb C}^*$, 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 \}.\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 $H = \{ 1, -1, i, -i \}$. 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.

##### Proof

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. \end{align*}