Suppose $f$ and $g$ are holomorphic in a region containing the disc $|z|\leq 1$. Suppose that $f$ has a simple zero at $z=0$ and vanishes nowhere else in $|z|\leq 1$. Let $$f_{\epsilon}(z)=f(z)+\epsilon g(z).$$Show that if $\epsilon$ is sufficiently small, then
(a) $f_{\epsilon}(z)$ has a unique zero in $|z|\leq 1,$ and
(b) if $z_{\epsilon}$ is this zero, the mapping $\epsilon\to z_{\epsilon}$ is continuous.
I have been asked to solve the question as homework.
Part (a) is actually trivial by applying Rouche's Theorem since $f$ and $g$ are holomorphic and we have $$|f(z)|>\epsilon|g(z)|$$ when $\epsilon$ is sufficiently small. But Part (b) is difficult for me. As the exercise show, the zero $z_{\epsilon}$ is a function of variable $\epsilon$, and I don't know how to determine where is the zero $z_{\epsilon}$. I can never show $$\lim_{h\to0}(z_{\epsilon+h}-z_{\epsilon})=0$$without knowing the exact expression of $z_{\epsilon}$.
So I decide to consider the inverse map: $z_{\epsilon}\to\epsilon$. And I get $$\epsilon=-\frac {f(z_{\epsilon})}{g(z_{\epsilon})}.$$ This is a continuous function. Can I end proof by saying "Since the inverse function is continuous"? Or my consideration is actually wrong? HELP! Please show me the correct proof or the right way solving the exercise.
