$\{(u,z)\in S^1\times D^2:u^cz^d=w\}$ is $S^1$ where $w\in D^2-\{0\}$ and $c,d$ are integers with $0\leq c<d$ and $(c,d)=1$

Question

According to Definition 16 of https://www.intlpress.com/site/pub/files/_fulltext/journals/pamq/2008/0004/0002/PAMQ-2008-0004-0002-a001.pdf, for a pair of integers $c,d$ with $0\leq c<d$ and $(c,d)=1$, the map $f:S^1\times D^2\to D^2$, $f(u,z)=u^cz^d$ is a $S^1$-bundle over $D^2-\{0\}$. I am trying to verify this. For $w\in D^2-\{0\}$, we have $f^{-1}(w)=\{(u,z):u^cz^d=w\}$. How do we know that this is a circle? Since $|w|=|u^cz^d|=|z|^d$, $|z|=|w|^{1/d}$ so in fact $f^{-1}(w)$ lies on a torus, but I can't see why it is a circle.