Why are Hopf manifolds compact?

Question

For a fixed $\lambda>0$, we define a Hopf manifold by the quotient of the group action $\mathbb Z \times \mathbb C \setminus \{0\} \to \mathbb C \setminus \{0\}, (n,z)\mapsto \lambda^n z$.
Is there a simple proof which shows that this manifold is compact?

When showing that other manifolds which are quotients, like projective spaces or tori, are compact, the easiest way is to show that the quotient map $\pi$ factors through a compact set.
But I don't see if this is the case.

Answer 1

Show that the quotient is the image of $\{z\in\Bbb C:1\le|z|\le\lambda\}$ which is a compact set (if $\lambda>1$). If $0<\lambda<1$ consider $1/\lambda$ instead, but if $\lambda=1$ the quotient isn't compact.

Answer 2

This manifold is the quotient of $\mathbb{C}^n-\{0\}$ by $x\rightarrow \lambda^nx$, where $|\lambda|<1$. The quotient map here factors through $S^{2n-1}$.