Each point $(x, y)$ on the unit circle can be represented as a complex number, $x + iy = \cos \theta + i \sin \theta$

Question

Show that the set of all complex numbers on the unit circle form a group under multiplication of complex numbers. Some helpful trig identities:

$$\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \sin (\alpha − \beta) = \sin \alpha \cos \beta − \cos \alpha \sin \beta \\ \cos (\alpha + \beta) = \cos \alpha \cos \beta − \sin \alpha \sin \beta \\ \cos (\alpha − \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$$

Can some one help me where to start, what I did was I know The set of all points on the unit circle are the complex numbers whose absolute value is $1$ and where $\theta$ is the angle between the positive $x$-axis and the line segment joining the origin and the point, so plugged $1$ into $1+i1= \cos \theta + i \sin \theta$. Then I don’t know what to do next

Answer 1

Let's denote the set of such complex numbers by $S^1.$

1.Choose any two elements from $S^1$ namely $a, b.$

Let $a= \cos\theta_1 +i \sin \theta_1$

and $b= \cos\theta_2 +i \sin \theta_2.$

Then $ab= (\cos\theta_1 +i \sin \theta_1)(\cos\theta_2 +i \sin \theta_2) = \cos(\theta_1+\theta_2) +i \sin (\theta_1 +\theta_2).$ Note that $\cos(\theta_1+\theta_2) +i \sin (\theta_1 +\theta_2)$ is in $S^1.$

2. Note that $1=\cos0 +i\sin0$ works as the identity element.

3. For any element $\cos\theta +i \sin \theta$ in $S^1,$ we have that ($\cos\theta +i \sin \theta$)($\cos(-\theta) +i \sin (-\theta)$)= 1. That means ($\cos(-\theta) +i \sin (-\theta)$) is the inverse of ($\cos\theta +i \sin \theta$).

4. Associativity is a result of pathetic computation.

If you know that $\cos\theta+ i\sin\theta = e^{i\theta},$ life will be much easier.

Answer 2

The circle group. By Euler's formula any such number is $e^{i\theta}=\cos\theta+i\sin\theta$.

Note that every point on the unit circle can indeed be written as $e^{i\theta}$, where $\theta$ is the angle with the $x$-axis.

Next the group operation is discovered by looking at $e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2)}$. So we just add the angles.

Answer 3

Use the Euler's formula; unit ciircle is given by $S=\{e^{i\theta}|\theta\in \mathbb{R}\}. $ Now

Closure law: for an two elements $e^{i\theta_1}, e^{i\theta_2}$ in S, $e^{i\theta_1}. e^{i\theta_2}=e^{i(\theta_1+\theta_2)}\in S. $

Similarly associativity can be established.

Identity law: $1=e^{i.0}$ is the Identity element.

Inverse law: For each $e^{i\theta}\in S$ there is $e^{i(-\theta)}\in S$ such that $e^{i\theta} e^{i(-\theta)}= e^{i(-\theta)} e^{i(\theta)} =e^0=1.$
Check the commutativity! Thus it forms an abelian group under multiplication.