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Section4.2Multiplicative Group of Complex Numbers

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*}

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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 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.

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Figure4.18Polar 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).\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 \(zw = 6 \cis( \pi / 2 ) = 6i\).

Proof
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

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*}

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Figure4.278th roots of unity