12: The Circle Group
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Before defining the circle group we first discuss some geometric aspects of the field of complex numbers. A typical element \(z\) of \(\mathbb{C}\) will be written \(z = x + yi\) where \(s, y \in \mathbb{R}\). We identify \(z = x+yi\) with the point \((x,y)\) in the plane. Thus the absolute value \(|z|\) of \(z\) is defined by \[|z| = \sqrt{x^2 + y^2}.\] Note that since \(z\overline{z} = x^2 + y^2\) we also have: \[|z| = \sqrt{z\overline{z}}.\]
Problem 12.1 Prove that for \(z,w \in \mathbb{C}\)
- \(|zw| = |z||w|\),
- \(|z| \ge 0\), and
- \(|z| = 0 \Longleftrightarrow z = 0\).
We know from analytic geometry that \(|z|\) represents the distance from \(z\) to the origin \(0\) in the plane. The directed angle \(\theta\) that the segment from \(0\) to \(z\) makes with the positive side of the \(x\)-axis is called the argument or polar angle of \(z\). As in polar coordinates we write \(r = |z|\). Then we have \[x = r \cos \theta,\] \[y = r \sin \theta,\] and \[\begin{align} \label{polarform} z = r (\cos\theta + i \sin\theta)\end{align}\] From trigonometry we know that every non-zero complex number \(z\) may be written uniquely in the form (12.5) for real numbers \(r\) and \(\theta\) satisfying \(r > 0\) and \(0 \le \theta < 2\pi\).
We assume that students are familiar with the exponential function \(x~\mapsto~e^x\) where \(x \in \mathbb{R}\). We extend the definition of this function from \(\mathbb{R}\) to \(\mathbb{C}\).
For \(z \in \mathbb{C}\) let \(z = x + yi\) where \(x,y \in \mathbb{R}\), We define the exponential function \(z \mapsto e^z\) by \[e^z = e^{x+yi} = e^x(\cos y + i \sin y.)\] in particular, if \(\theta \in \mathbb{R}\) we have \[e^{i\theta} = \cos \theta + i \sin \theta .\]
From the above we have immediately the following:
Every non-zero complex number \(z\) may be written uniquely in the form \[\begin{aligned} z = re^{i\theta}\end{aligned}\] where \(r = |z| > 0\) and \(0 \le \theta < 2\pi\). \(\blacksquare\)
Note that the expression \(e^{i\theta}\) is well-defined for all \(\theta \in \mathbb{R}\).
Let \(z_1 = r_1 e^{i\theta_1}\) and \(z_2 = r_2 e^{i\theta_2}\) where \(r_i \ge 0\) and \(\theta_i\) are real numbers. Then \[z_1z_2 = r_1r_2e^{i(\theta_1 + \theta_2)}. \rule{6pt}{6pt}\]
Problem 12.2 Use the addition identities for the sine and cosine to prove Theorem 12.2.
Note that, in words, Theorem 12.2 says: The argument of the product is the sum of the arguments of the factors and the absolute value of the product is the product of the absolute values of the factors.. This easily generalizes via induction to the following: If \(z_j= r_je^{i \theta_j}\), \(j = 1, \dots, n\) are complex numbers then \[z_1z_2\cdots z_n = r_1r_2\cdots r_n e^{(i\theta_1 + \theta_2 + \cdots + \theta_n)}.\] Taking \(r_j = 1\) for all \(j\) we obtain the following famous theorem:
For all \(\theta \in \mathbb{R}\) and \(n \in \mathbb{Z}\), we have \[(\cos (\theta) + i\sin (\theta) )^n = \cos(n\theta) + i\sin(n\theta),\] equivalently, \[(e^{i\theta})^n = e^{in\theta}. \rule{6pt}{6pt} \blacksquare\]
We define \[\mathbb{T}= \{ z \in \mathbb{C}\, | \, |z| = 1 \};\] \(\mathbb{T}\) is a group with respect to multiplication in \(\mathbb{C}\) and is called the circle group.
Note that geometrically \(\mathbb{T}\) is the set of complex numbers which are at a distance 1 from the origin, that is, it’s points are exactly the points on the unit circle \(x^2 + y^2 = 1\).
Problem 12.3 Show that every element \(z \in \mathbb{T}\) may be uniquely written in the form \(z = e^{i\theta}\) where \(0 \le \theta< 2\pi\).
Problem 12.4 Prove that \(\mathbb{T}\) is a subgroup of \(U(\mathbb{C})\).
Problem 12.5
(a) Prove that the mapping \(\varphi:\mathbb{T}\to\mathbb{C}\) defined by \(\varphi(\theta) = e^{i\theta}\) is a homomorphism from \((\mathbb{R},+)\) onto the circle group \(\mathbb{T}\).
(b) Show that for every point \(z \in \mathbb{T}\) there are infinitly many \(\theta\in \mathbb{R}\) such that \(\varphi(\theta) = z\).
Recall that in Problem 10.15 we showed that complex numbers can be represented as certain \(2 \times 2\) matrices over the real numbers. So it should come as no surprise that the circle groups can also be represented by certain \(2 \times 2\) matrices over the real numbers. It turns out that this set of matrices also has another name which we give in the following definition.
Define \[SO(2) = \left \{ \left ( \begin{array}{cr} \cos\theta& -\sin\theta\\ \sin\theta&\cos\theta \end{array} \right ) \ | \ \theta\in \mathbb{R} \right \}.\] \(SO(2)\)is a subgroup of \(SL(2,\mathbb{R})\) and is called the special orthogonal group of degree 2.
For \(\theta\in \mathbb{R}\), define \[R(\theta) = \left ( \begin{array}{cr} \cos\theta& -\sin\theta\\ \sin\theta&\cos\theta \end{array} \right )\]
With this definition we have \(SO(2,\mathbb{R}) = \{ R(\theta) \, | \, \theta\in \mathbb{R} \}\).
Problem 12.6 Prove (a) that \(R(\theta_1)R(\theta_2) = R(\theta_1+\theta_2)\), (b) \(R(0)\) is the \(2 \times 2\) identity matrix, and (c) \(R(\theta)^{-1} = R(-\theta)\). Conclude that \(SO(2,\mathbb{R})\) is a subgroup of \(GL(2,\mathbb{R})\).
Problem 12.7 Prove that \(SO(2,\mathbb{R}) \cong \mathbb{T}\).
Problem 12.8 Prove that if we represent a point \(p=(x,y)\) in the plane by a \(2\times 1\) matrix \(\left[ \begin{array}{c} x \\ y \end{array} \right ]\) then the point \(R(\theta) p\) given by the matrix product \[R(\theta)p = \left [ \begin{array}{cr} \cos\theta& -\sin\theta\\ \sin\theta&\cos\theta\end{array} \right ] \left[ \begin{array}{c} x \\ y \end{array} \right]\] is obtained by rotating \(p\) through \(\theta\) radians counter-clockwise about the origin. [Hint use the polar coordinate representation \((x,y) = (r\cos \theta, r \sin \theta)\) of the point \(p\).]
The above problem also justifies referring to the circle group as the group of rotations of the plane.
We now determine the order of an element \(e^{i\theta} \in \mathbb{T}\).
An element \(z = e^{i\theta} \in \mathbb{T}\) has finite order if and only if \(\theta= \frac k n \pi\) for some \(n \in N\) and \(k \in \mathbb{Z}\), that is, if and only if \(\theta\) is a rational multiple of \(\pi\).
Proof First we recall from trignometry that \((\cos \alpha, \sin \alpha) = (1,0)\) if and only if \(\alpha = 2\pi k\) for some integer \(k\). Using exponentional notation, this says that \(e^{i\alpha} = 1\) if and only if \(\alpha = 2\pi k\) for some integer \(k\).
Assume that \(e^{i\theta}\) has finite order. Then by De Moivre’s Theorem we have \(e^{in\theta} = 1\) and by the previous remark, \(n \theta= 2 \pi k\) for some integer \(k\). Solving for \(\theta\) we see that \(\theta= \frac {2k}{n} \pi = \frac {k'}{n} \pi\) where \(k' = 2k\). That is, \(\theta\) is a rational multiple of \(\pi\). Conversely, suppose that \(\theta= \frac k n \pi\) for some \(n \in N\) and \(k \in \mathbb{Z}\). Then \[(e^{i\theta})^{2n} = e^{i(\theta{2n})} = e^{i \frac k n 2n \pi} = e^{ik2\pi} = 1.\] This shows that the order of \(e^{i\theta}\) is finite and at most \(2n\). \(\blacksquare \)
Problem 12.9 Show that the order of the element \(e^{i\sqrt{2}\pi}\) in \(\mathbb{T}\) is infinite. What about the element \(e^{i\sqrt{2}}\)? (For the latter you may assume that \(\pi\) is transcendental.)
Let \(n \in \mathbb{N}\). An element \(z \in \mathbb{C}\) is said to be an \(n\)-th root of unity if \(z^n = 1\).
Problem 12.10 Prove that for \(n \in \mathbb{N}\) the set \[\begin{align} \label{roots_of_unity} \{z \in \mathbb{C}\, | \, z^n = 1 \}\end{align}\] is a subgroup of \(U(\mathbb{C})\).
The set (12.17) of all \(n\)-th roots of unity is a subgroup of \(U(\mathbb{C})\) called the group of \(n\)-th roots of unity.
Figure 12.1: The 12th roots of unity (= the vertices of the regular 12-gon).
Problem 12.11 Prove that \(z \in \mathbb{C}\) is an \(n\)-th root of unity if and only if \(z\) is an element in \(\mathbb{T}\) of finite order \(k\) where \(k \, | \, n\).
For \(n \in \mathbb{N}\) define \[\zeta_n = e^{i\frac {2 \pi}n}.\]
The group of \(n\)-th roots of unity is cyclic of order \(n\). One generator of the group is \(\zeta_n\)
Proof From De Moivre’s Theorem it is clear that \((\zeta_n)^n = 1\). Note that the powers \[(\zeta_n)^k = e^{i k \frac {2 \pi}n}, \quad k = 0, 1, \dots , n-1\] are the vertices of the regular \(n\)-gon centered at the origin. Hence \((\zeta_n)^k \neq 1\) for \(0 < k < n\). This proves that \(o(\zeta_n) = n\).
Now, suppose that \(z\) is any \(n\)-th root of unity. Note that \(|z|^n = |z^n| =1\). That is, \(|z|\) is a positive real number whose \(n\)-th power is 1. It follows that \(|z|\) must be equal to 1. Hence \(z = e^{i\theta}\). By the argument in the proof of Theorem 12.4 since \(z^n = 1\), we have \(\theta= k\frac {2 \pi}n\). This shows that \(z = e^{i k \frac {2 \pi}n} = (\zeta_n)^k\), and therefore lies in the subgroup \(\langle\,\zeta_n\,\rangle\) generated by \(\zeta_n\). \(\blacksquare\)
Problem 12.12 Show that \(z \in \mathbb{T}\) if and only if \(z^{-1} = \overline{z}\).
Problem 12.13 Show that if \(z = e^{i\theta}\) then \(\overline{z} = e^{-i \theta}\).
Problem 12.14 Use the formula for \(R(\theta)\) to find the coordinates of the point \((1,1) \in \mathbb{R}^2\) after it has been rotated \(30^o\) counter-clockwise about the origin. Do the same for \(60^o\). Express the coordinates of the answer as rational numbers and/or radicals, not trig functions.
Problem 12.15 Prove that the group \(\langle \, \zeta_n \, \rangle\) is isomorphic to the group \(\mathbb{Z}_n\) under addition modulo \(n\).
Problem 12.16 For each \(n \in\{1,2,3,4,6,8 \}\) find all the \(n\)-th roots of unity \((\zeta_n)^k\) for \(k \in \{0,1,\cdots, n-1\}\). Express them in the form \(a + bi\) where \(a\) and \(b\) are real numbers not involving trig functions. Also sketch the location in the plane of the \(n\)-roots of unity for each \(n\).
Problem 12.17 Prove that \(\langle \, e^{i\pi\sqrt{2}} \, \rangle \cong \mathbb{Z}\).