Determining if something is an invariant I'm stuck trying to show that this is true.
Let $G$ be a group that acts on a set $X$.  Let $Y$ be any set.  A function $f: X \to Y$ is said to be invariant if given any $x_1,x_2 \in X$ with $x_1, x_2$ on the same orbit, then $f(x_1)=f(x_2)$.  
Let $G$ be the unit circle in $\mathbb{C}^*$ (Complex numbers under multiplication) and $X=\mathbb{C}$ where $G$ acts on $X$ by multiplication.  Let $f: X \to \mathbb{R}$ by given by the rule $f(z)=|z|$.
(a)  Show that this is an invariant.
First, we have to figure out what the orbits are.  I think the orbits in this group action are "small" in the sense that they only contain one element.  So if two elements shared an orbit, they would be the same point.  Hence we get for free that $f(z_1)=f(z_2)$.
 A: First of all, let $\alpha:(g,z)\in G\times \mathbb{C}\mapsto\alpha(g,z)\in\mathbb{C}$ be the given action (maybe the definition of action you have is different: the action may be something like $\alpha:G\rightarrow Bij(\mathbb{C})=\left\{bijections\ on\ \mathbb{C}\right\}$. In this case, you should read $\alpha(g,z)=\alpha(g)(z)$).
By definition, the action $\alpha$ is given by $\alpha(g,z)=gz$, for every $g\in G$ and $z\in\mathbb{C}$. Let $G(z)=\left\{\alpha(g,z)=gz:g\in G\right\}$ denote the orbit of an element $z\in\mathbb{C}$.
Fix $z\in\mathbb{C}$.

Let's show that $G(z)=\left\{w\in\mathbb{C}:|w|=|z|\right\}$.

If $w\in G(z)$, then $w=gz$ for some $g\in G$, so $|w|=|gz|=|g||z|=1|z|=|z|$.
On the other hand, if $|w|=|z|$, define $p\in\mathbb{C}$ by
$$p=\begin{cases}{w}/{z}&\text{, if }z\neq 0\\
1&\text{, if }z=0.\end{cases}$$
In both cases, $|p|=1$, so $p\in G$, and $w=pz\in G(z)$.
Now it is easy to show that $f$ is invariant. If $z_1$ and $z_2$ are elements of $\mathbb{C}$ which belong to the same orbit, say $G(w)$, then, by the above assertion,
$f(z_1)=|z_1|=|w|=|z_2|=f(z_2)$. Hence, $f$ is invariant.
