Show that $S^1$ acts on $S^3$

Question

$S^3=\{(z_1, z_2) \in \mathbb{C^2} \mid |z_1|^2 + |z_2|^2 = 1 \}$

Show that $S^1$ acts on $S^3$ by $z \cdot (z_1, z_2)=(zz_1, zz_2)$

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An action of a topological group $G$ on a topological space $X$ (by homeomorphisms) is a continuous map $F : G \times X \rightarrow X$ such that the following $3$ conditions hold:

1. For each fixed $g \in G$ the map $F(g, \cdot) : X \rightarrow X$ is a homeomorphism
2. The identity element acts as identity map
3. Product is respected: $F(gh, a)=F(g, F(h, a))$

So in this case, I think $G=S^1$ and $X=S^3$

1. Is $z \cdot (z_1, z_2)=(zz_1, zz_2)$ a homeomorphism? I believe it is due to continuity and bijectivity... but I do not know about an inverse? Could someone help me with that?
2. $e \in S^1$ gives $e \cdot (z_1, z_2)=(ez_1, ez_2)$ so I think this works...
3. And products seem to be respected

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I am not sure if I am misunderstanding these criteria, if someone could help me outline this in more detail that would be great