Let $f$ be a holomorphic function. Suppose $f(z)$ has a zero of order $n$ at $0$. Show that for some non-zero sufficiently small $a$, the equation $f(z) = a$ has $n$ solutions of multiplicity one.
I know that for a neighborhood $U$ of $0$ we can write $f(z) = z^ng(z)$. I want to show that we can find a function $h$ such that $h(z)^n=g(z)$. But I am not sure how to proceed. Any idea?
Since $g$ is continuous and $g(0)\ne 0$, you can choose $U$ small enough that $g(U)$ fits within a ball that doesn't contain $0$. On that ball you can choose a single branch of the $n$th-root function and compose it with $g$.
You can use the argument principle. $f(z)$ and $f(z)-a$ have the same derivative $f'(z)$, and they have the same number of zeroes (with multiplicities) on $U$.
Now, $f'$ is holomorphic, so its zeroes are isolated. If $U$ is chosen so that $f'(z)=0$ only on $z=0$, then all the zeroes of $f(z)-a$ must be simple.