The set of complex numbers of modulus $1$ is a group under multiplication Show that $C=\{z\in \mathbb{C} \mid  |z|=1\}$ is a group under complex multiplication.
I'm a little confused because isn't the identity the only element with order $1$?  What is this set?
 A: We ﬁrst show  that $G = \mathbb{C}^* = \mathbb{C} − \{0\}$ under complex multiplication forms a group.


*

*Closure: 


Let $z = a + bi$ and $w = c + di$ be both in $\mathbb{C}^∗$. Then we have
$zw = (a + bi)(c + di) = ac − bd + (ad + bc)i$. To show that this element is in $G$, we
compute $$(ac − bd)^2 + (ad + bc)^2 = a^2c^2 − 2abcd + b^2 d^2 + a^2 d^2 + 2abcd + b^2c^2
\\= a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2
\\= a^2(c^2 + d^2) + b^2(c^2 + d^2)
\\= (a^2 + b^2)(c^2 + d^2).$$
Now $z, w ∈ G$ imply that a $a^2+b^2 > 0$ and $c^2+d^2 > 0$, whence $(ac−bd)^2+(ad+bc)^2 > 0$ and $zw ∈ G$.


*

*Associativity:


To show associativity, let $z = a + bi, w = c + di$ and $u = e + fi$. Then
$$(zw)u = [(a + bi)(c + di)](e + f i)
= [(ac − bd) + (ad + bc)i](e + f i)
\\= [(ac − bd)e − (ad + bc)f] + [(ac + bd)f + (ad + bc)e]i
\\= (ace − bde − adf − bcf) + (acf − bdf + ade + bce)i
\\= [(a(ce − df) − b(cf + de)] + [a(cf + de) + b(ce − df)]i
\\= (a + bi)[(ce − df) + (cf + de)i]
\\= (a + bi)[(c + di)(e + f i)]
\\= z(wu).$$


*

*Identity Element and Inverse:


It is not diﬃcult to check that $1 + 0i$ acts as a unit in multiplication and that 
$$(a + bi)^{−1} =
\frac{1}{a+bi} =\frac{a−bi}{(a+bi)(a−bi)} = \frac{a−bi}{a^2+b^2} = \frac{a}{a^2+b^2} −
\frac{b}{a^2+b^2} i ∈ G$$
, since 
$$\left(\frac{a}{a^2+b^2}\right)^2 + \left(\frac{b}{a^2+b^2}\right)^2 = \left(\frac{a^2+b^2}{(a^2+b^2)^2}\right) = \frac{1}{a^2+b^2} > 0$$
Since all the group axioms hold, $(G,\cdot)$, is a group under multiplication.
Ok, this is a group, now we let see that Circle Group is a Group.
Let $K$ be the set of all complex numbers of unit modulus:
$K={z∈\mathbb{C}:|z|=1}$


*

*Closure:


Then the circle group $(K,\cdot)$ is an uncountably infinite abelian group under the operation of complex multiplication.
$$
   z,w  ∈   K                      
⟹   |z|  =   1=|w|                      
⟹   |zw|  =   |z||w|                      
⟹   zw  ∈   K $$
So $(S,\cdot)$ is closed.


*

*Associativity, comes from complex multiplication is associative.

*Identity : From Complex multiplication identity is one we have that the identity element of $K$ is $1+0i$.

*Inverses
We have that $|z|=1⟹\frac{1}{|z|}=\left|\frac{1}{z}\right|=1$.
But $z\cdot\frac{1}{z}=1+0i$.
So the inverse of $z$ is $\frac{1}{z}$.


*

*Commutative: We have that Complex Multiplication is Commutative


So $K$ is a subgroup of $G$ under complex multiplication.
We can assert that from complex multiplication is commutative it also follows from Subgroup of Abelian Group is Abelian that $K$ is an abelian group.
A: Hint: prove that if you multiply two unitary complex numbers then the result is also an unitary number.  
I suggest you to learn about polar representation of a complex number. This could make the solution easier.
A: Regarding your second question: in $\mathbb{C}$, you have a lot of elements with modulus $1$ — all $e^{i\theta}$ for $\theta\in[0,2\pi)$.
A: Every complex numbers could be written as $re^{i\theta}$, where $r\ge0$ and $\theta\in[0,2\pi[$. In particular $r$ is the modulus of that complex number, i.e. $r=|re^{i\theta}|$. So it will be $r=1$. Hence 
$$
C=\{z\in\mathbb C\;:\;|z|=1\}=\{e^{i\theta}\;:\;\theta\in[0,2\pi[\}
$$
which is a group under multiplication, since


*

*Contains a unity which is $1=e^{i\cdot0}$

*Every element admits inverse $(e^{i\theta})^{-1}=e^{i(-\theta)}$ which is already in $C$.

*It's closed under mutiplication: $e^{i\theta}e^{i\psi}=e^{i(\theta+\psi)}$ and moreover the multiplication is associative.

