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Section 4.2 Multiplicative 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 \(|z| = \sqrt{a^2 + b^2}\text{.}\)

Example 4.16.

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

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


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


\begin{align*} z^{-1} & = \frac{2}{13} - \frac{3}{13}i\\ |z| & = \sqrt{13}\\ \overline{z} & = 2-3i\text{.} \end{align*}
The complex plane where the horizontal axis is the x-axis or real axis and the verticle axis is the y-axis or imaginary axis. The point z1 = 2 + 3i is in the upper right quadrant, the point z2 = 1- 2i in the lower right quadrant, and z3 = -3 + 2i in the upper right quadrant.
Figure 4.17. Rectangular 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 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.

The complex plane where the horizontal axis is the x-axis or real axis and the verticle axis is the y-axis or imaginary axis.  The point a + bi is in the upper right quadrant.  The point is also determined by a ray that at an angle of theta counterclockwise from the horizontal axis having a length of r.
Figure 4.18. Polar coordinates of a complex number

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*}


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


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

Example 4.19.

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

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


\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*}


\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.

Example 4.21.

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*}

Example 4.23.

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*}

Subsection 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 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.

Example 4.26.

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*}
The 8 roots of unity are spaced evenly around the unit circle beginning with 1 on the positive horizontal axis and followed by omega, i, the cube of omega, -1, omega to the fifth power, -i, and omega to the seventh power.
Figure 4.27. 8th roots of unity