Let f, g: $S^1 \rightarrow S^1$ be continuous mappings and let's take the complex multiplication operation on $S^1 \subset \mathbb{C}$. Define $h(z)$ to be the product $h(z)=f(z)* g(z)$. Show that $deg(h)=deg(f)+deg(g)$
I'm having a difficult time understanding what the problem is referring when it asks to "take the complex multiplication operation on $S^1 \subset \mathbb{C}$"
I would appreciate some help getting started on this problem as well, thank you.
You are wanting to consider the elements in $\Bbb C$ with norm $1$, i.e. $$\Bbb S^1=\{z\in\Bbb C\mid |z|=1\}=U(1),$$
I.e. you are taking $e^{i\theta_1}\times e^{i\theta_2}=e^{i(\theta_1+\theta_2)}$ - the standard multiplication of these complex numbers. So where $f,g:\Bbb S^1\to \Bbb S^1$, you are just computing the multiplication point-wise $h=f\times g$ means $h(z)=f(z)g(z)$.
The group $U(1)$ above is the group of unitary $1$ by $1$ matrices over $\Bbb C$. $A\in U(n)$ if $A$ is an $n$ by $n$ matrix, and $AA^*=I$, where $A^*$ is the conjugate transpose of $A$. For $U(1)$ these are just $1$ by $1$ matrices with $$AA^*=[z][\bar{z}]=[z\bar{z}]=[|z|^2]=I_{1\times 1}=1.$$