Composition of holomorphic functions and order of zeros Consider the holomorphic functions $f_1 \in H(\Omega_1)$ and $f_2 \in H(\Omega_2)$, where $\Omega_1, \Omega_2 \subset \mathbb{C}$ and $f_1(\Omega_1) \subset \Omega_2$. Let $$g = f_2 \circ f_1.$$ 
Consider $z_0 \in \Omega_1$ and $z'_0 = f_1(z_0) \in \Omega_2$. 
Suppose $z'_0$ is a zero of order $k$ for $f_2$. Is $z_0$ a zero of order $k$ for $g$? If not what is it? Do we need to deal with the case where the derivatives of $f_1$ vanish separately?
 A: Assuming $f_1$ is nonconstant, $g$ has a zero of order at least $k$ at $z_0$.  It is exactly $k$ if and only if $f_1'(z_0)\neq 0$, which is equivalent to $f_1(z)-z_0'$ having a zero of order $1$ at $z_0$.  
In general, if $f_1(z)-z_0'$ has a zero of order $m$ at $z_0$, then $g$ has a zero of order $km$ at $z_0$.  To see this, note that $f_1$ and $f_2$ have local representations $f_1(z)=z_0'+(z-z_0)^m g_1(z)$ near $z_0$ and $f_2(z)=(z-z_0')^kg_2(z)$ near $z_0'$, with $g_1(z_0)\neq 0$ and $g_2(z_0')\neq 0$.  Then $$g(z)=f_2(f_1(z))=(z_0'+(z-z_0)^mg_1(z)-z_0')^kg_2(f_1(z))=(z-z_0)^{mk}(g_1(z))^kg_2(f_1(z)).$$
Because $(g_1(z_0))^kg_2(f_1(z_0))\neq 0$, this shows that $g$ has a zero of multiplicity $mk$ at $z_0$.
A: Another point of view (ultimately equivalent to @JonasMeyer's answer).
Without loss of generality, $z_0 = z_0' = 0$. Say $f_1$ has order $m$ at $z_0$, then we can write the two functions as
\begin{align}
f_1(z) =&\ \sum_{n_1\ge m}a_{n_1}z^{n_1},\\
f_2(z) =&\ \sum_{n_2\ge k}b_{n_2}z^{n_2}
\end{align}
in a neighborhood of $0$, with $a_m,b_k\neq0$. Their composite then has the form
$$g(z) = a_mb_kz^{mk} + \sum_{n>mk}c_nz^n.$$
Therefore, $g$ has a zero of order $mk\ge k$ at $z_0$.
