sum of multiplicities of the pre-image points for the map $z\mapsto z^m$

sum of multiplicities of the pre-image points for the map $f:D\rightarrow D,z\mapsto z^m$ is m? but I dont understand why, well, $0$ has multiplicity $m$ and all other $(m-1)$ points which are $m'th$ roots of some $\omega$(say) has multiplicity exactly 1, so the sum will be $m+(m-1)=2m-1$? $D$ open unit disk.Thank you.

-
Why are you summing up the multiplicities at the pre-images of different points? Just pick one $w\in D$ and consider $f^{-1}(w)$ alone. Also, if $w\neq0$ then $f^{-1}(w)$ consists of $m$ points (the $m$-th roots of $w$), not $m-1$. –  Andrea Mori Jul 26 '12 at 9:07
aha aha aha, I see.Thank you @AndreaMori for example here I have summed up multiplicities of $0$ and $\omega$ right? –  Bunuelian Trick Jul 26 '12 at 9:08
you did that, yes. :) –  Andrea Mori Jul 26 '12 at 9:12