If the analytic function $f(z)$ has a zero of order $N$ at $z_0$ then $f(z)=g(z)^N$

Question

show that if the analytic function $f(z)$ has a zero of order $N$ at $z_0$ then $f(z)=g(z)^N$ for some function $g(z)$ analytic near $z_0$ and satisfies $g'(z_0) \neq 0$

what i tried

here $f(z)$ has a zero of order $N$ at $z_0$

so we can write $f(z)=(z-z_0)^Nh(z)$

where $h(z_0)\neq 0$

but how to prove our conclusion

Answer 1

Write $f(z)=a (z-z_0)^N f_1(z)$ with $f_1(z_0)=1$ and $a\not=0$. Then $f_1(z)=1+f_2(z)$ where $f_2$ is analytic and satisfies $f_2(z_0)=0$. Thus the function $g_1$, $g_1(z):=\sum_{j=0}^\infty \binom{1/N}{j} f_2(z)^j$ is analytic near $z_0$ and $g_1(z)^N=f_1(z)$. Therefore $g:=\sqrt[N]{a} g_1$ satisfies the condition.

Answer 2

Taking $g(z)=\left(z-z_{0}\right) e^{Log(h(z)/ N}$ for an appropriate branch of logarithm \begin{equation*} g(z)^N=\left(z-z_{0}\right)^N e^{Log(h(z))} = \left(z-z_{0}\right)^N h(z) = f(z) \end{equation*} with $h(z_0) \neq 0$ and is easy to check $g'(z_0) \neq 0$.