Denote $\mathbb{D}_r^1\subset\mathbb{C}$ to the open disk of radius $r>0$, centered at the origin, and $\mathbb{D}_r^n=\prod_{j=1}^n\mathbb{D}^1_r$ to the $n$-polydisk of polyradius $(r,r,\dots,r)$.
I want to prove the following result, that showed up as a step on a proof from a several complex variables course I'm following:
Lemma. Let $G=G'\times G_n\subset\mathbb{C}^n$ be a domain (open and connected subset) that contains the origin, where $G'\subset\mathbb{C}^{n-1}$ and $G_n\subset \mathbb{C}$, and let $f:G\to\mathbb{C}$ be holomorphic and non-zero. Suppose that $z_n\mapsto f(0',z_n)$ is non-zero and has a root at zero of multiplicity $k$. Then there are $\varepsilon,\delta>0$ such that $\mathbb{D}_\delta^{n-1}\times\mathbb{D}_\varepsilon^1\subset G$, and for each $|z'|<\delta$, the one-variable holomorphic function $z_n\in \mathbb{D}_\varepsilon^1\mapsto f(z',z_n)$ has $k$ roots (counted with multiplicity).
On the lecture it was argued that this result "followed from Rouché's theorem," but no additional details were supplied. I am not sure how to apply Rouché's theorem as it is stated in wikipedia, since I don't know how $f(0',-)$ and $f(z',-)$ behave at $\partial\mathbb{D}_\varepsilon^1$.
Another possibility I thought of is Hurwitz's theorem, but this theorem deals with sequences of holomorphic functions and yields a different $\delta>0$ for each sequence of values of $z'$ converging to zero, so from here I don't see how to get a universal $\delta>0$ that makes the result work for all $|z'|<\delta$.
Okay so the proof I devised is similar to the proof of Hurwitz's theorem, but we'll replace uniform convergence on compact sets for uniform continuity.
Choose some $\varepsilon>0$ such that $\mathbb{D}_\varepsilon^1\subset G_n$ and $f(0',z_n)\neq 0$ for all $0<|z_n|\leq\varepsilon$. Such an $\varepsilon$ exists since the function $z_n\mapsto f(0',z_n)$ is holomorphic and non-zero. Choose $\lambda>0$ such that $|f(0',z_n)|>\lambda$ for all $|z_n|=\varepsilon$. Pick $r>0$ such that $\overline{\mathbb{D}}^{n-1}_r\subset G'$. Since $f(z',z_n)$ is uniformly continuous on the compact set $K(r,\varepsilon):=\overline{\mathbb{D}}_r^{n-1}\times\partial\mathbb{D}_\varepsilon^1$, there is $0<\delta\leq r$ such that $|f(z',z_n)|>\lambda/2$ for all $(z',z_n)\in K(\delta,\varepsilon)$. Therefore, the function $$ g(z',z_n):=\frac{\partial_{z_n}f(z',z_n)}{f(z',z_n)} $$ is well-defined for all values $(z',z_n)\in K(\delta,\varepsilon)$ (the denominator does not vanish). Since $g$ is uniformly continuous on $K(\delta,\varepsilon)$ and taking a smaller $\delta$ if necessary, we can assume that $$ |g(z',z_n)-g(0',z_n)|<\frac{1}{2\varepsilon} $$ for all $(z',z_n)\in K(\delta,\varepsilon)$. Then, for $z'\in\overline{\mathbb{D}}^{n-1}_\delta$, we have
$$ \frac{1}{2\pi }\left|\int_{|\zeta_n|=\varepsilon}\frac{\partial_{z_n}f(z',\zeta_n)}{f(z',\zeta_n)}d\zeta_n -\int_{|\zeta_n|=\varepsilon}\frac{\partial_{z_n}f(0',\zeta_n)}{f(0',\zeta_n)}d\zeta_n\right|<\frac{1}{2}. $$
But the LHS of this inequality equals the difference of the number of zeroes, inside $\mathbb{D}_\varepsilon^1$, of $z_n\mapsto f(z',z_n)$ and of $z_n\mapsto f(0',z_n)$ (Argument principle). Therefore it is an integral value. Hence, it vanishes and the two functions have the same number of zeroes.