Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/fontdata.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

12: The Circle Group

( \newcommand{\kernel}{\mathrm{null}\,}\)

Before defining the circle group we first discuss some geometric aspects of the field of complex numbers. A typical element z of C will be written z=x+yi where s,yR. We identify z=x+yi with the point (x,y) in the plane. Thus the absolute value |z| of z is defined by |z|=x2+y2. Note that since z¯z=x2+y2 we also have: |z|=z¯z.

Problem 12.1 Prove that for z,wC

  1. |zw|=|z||w|,
  2. |z|0, and
  3. |z|=0z=0.

We know from analytic geometry that |z| represents the distance from z to the origin 0 in the plane. The directed angle θ 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=rcosθ, y=rsinθ, and z=r(cosθ+isinθ) 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 θ satisfying r>0 and 0θ<2π.

We assume that students are familiar with the exponential function x  ex where xR. We extend the definition of this function from R to C.

Definition 12.1:

For zC let z=x+yi where x,yR, We define the exponential function zez by ez=ex+yi=ex(cosy+isiny.) in particular, if θR we have eiθ=cosθ+isinθ.

From the above we have immediately the following:

Theorem 12.1

Every non-zero complex number z may be written uniquely in the form z=reiθ where r=|z|>0 and 0θ<2π.

Note that the expression eiθ is well-defined for all θR.

Theorem 12.2

Let z1=r1eiθ1 and z2=r2eiθ2 where ri0 and θi are real numbers. Then z1z2=r1r2ei(θ1+θ2).

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 zj=rjeiθj, j=1,,n are complex numbers then z1z2zn=r1r2rne(iθ1+θ2++θn). Taking rj=1 for all j we obtain the following famous theorem:

Theorem 12.3 (De Moivre's Theorem)

For all θR and nZ, we have (cos(θ)+isin(θ))n=cos(nθ)+isin(nθ), equivalently, (eiθ)n=einθ.

Definition 12.2:

We define T={zC||z|=1}; T is a group with respect to multiplication in C and is called the circle group.

Note that geometrically 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 x2+y2=1.

Problem 12.3 Show that every element zT may be uniquely written in the form z=eiθ where 0θ<2π.

Problem 12.4 Prove that T is a subgroup of U(C).

Problem 12.5

(a) Prove that the mapping φ:TC defined by φ(θ)=eiθ is a homomorphism from (R,+) onto the circle group T.

(b) Show that for every point zT there are infinitly many θR such that φ(θ)=z.

Recall that in Problem 10.15 we showed that complex numbers can be represented as certain 2×2 matrices over the real numbers. So it should come as no surprise that the circle groups can also be represented by certain 2×2 matrices over the real numbers. It turns out that this set of matrices also has another name which we give in the following definition.

Definition 12.3:

Define SO(2)={(cosθsinθsinθcosθ) | θR}. SO(2)is a subgroup of SL(2,R) and is called the special orthogonal group of degree 2.

Definition 12.4:

For θR, define R(θ)=(cosθsinθsinθcosθ)

With this definition we have SO(2,R)={R(θ)|θR}.

Problem 12.6 Prove (a) that R(θ1)R(θ2)=R(θ1+θ2), (b) R(0) is the 2×2 identity matrix, and (c) R(θ)1=R(θ). Conclude that SO(2,R) is a subgroup of GL(2,R).

Problem 12.7 Prove that SO(2,R)T.

Problem 12.8 Prove that if we represent a point p=(x,y) in the plane by a 2×1 matrix [xy] then the point R(θ)p given by the matrix product R(θ)p=[cosθsinθsinθcosθ][xy] is obtained by rotating p through θ radians counter-clockwise about the origin. [Hint use the polar coordinate representation (x,y)=(rcosθ,rsinθ) of the point p.]

Remark

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 eiθT.

Theorem 12.4

An element z=eiθT has finite order if and only if θ=knπ for some nN and kZ, that is, if and only if θ is a rational multiple of π.

Proof First we recall from trignometry that (cosα,sinα)=(1,0) if and only if α=2πk for some integer k. Using exponentional notation, this says that eiα=1 if and only if α=2πk for some integer k.

Assume that eiθ has finite order. Then by De Moivre’s Theorem we have einθ=1 and by the previous remark, nθ=2πk for some integer k. Solving for θ we see that θ=2knπ=knπ where k=2k. That is, θ is a rational multiple of π. Conversely, suppose that θ=knπ for some nN and kZ. Then (eiθ)2n=ei(θ2n)=eikn2nπ=eik2π=1. This shows that the order of eiθ is finite and at most 2n.

Problem 12.9 Show that the order of the element ei2π in T is infinite. What about the element ei2? (For the latter you may assume that π is transcendental.)

Definition 12.5:

Let nN. An element zC is said to be an n-th root of unity if zn=1.

Problem 12.10 Prove that for nN the set {zC|zn=1} is a subgroup of U(C).

Definition 12.6:

The set (12.17) of all n-th roots of unity is a subgroup of U(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 zC is an n-th root of unity if and only if z is an element in T of finite order k where k|n.

Definition 12.7:

For nN define ζn=ei2πn.

Theorem 12.5

The group of n-th roots of unity is cyclic of order n. One generator of the group is ζn

Proof From De Moivre’s Theorem it is clear that (ζn)n=1. Note that the powers (ζn)k=eik2πn,k=0,1,,n1 are the vertices of the regular n-gon centered at the origin. Hence (ζn)k1 for 0<k<n. This proves that o(ζn)=n.

Now, suppose that z is any n-th root of unity. Note that |z|n=|zn|=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=eiθ. By the argument in the proof of Theorem 12.4 since zn=1, we have θ=k2πn. This shows that z=eik2πn=(ζn)k, and therefore lies in the subgroup ζn generated by ζn.

Problem 12.12 Show that zT if and only if z1=¯z.

Problem 12.13 Show that if z=eiθ then ¯z=eiθ.

Problem 12.14 Use the formula for R(θ) to find the coordinates of the point (1,1)R2 after it has been rotated 30o counter-clockwise about the origin. Do the same for 60o. Express the coordinates of the answer as rational numbers and/or radicals, not trig functions.

Problem 12.15 Prove that the group ζn is isomorphic to the group Zn under addition modulo n.

Problem 12.16 For each n{1,2,3,4,6,8} find all the n-th roots of unity (ζn)k for k{0,1,,n1}. 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 eiπ2Z.


    This page titled 12: The Circle Group is shared under a not declared license and was authored, remixed, and/or curated by W. Edwin Clark via source content that was edited to the style and standards of the LibreTexts platform.

    Support Center

    How can we help?