My question is one from complex analysis.Suppose $f(z)$ is a holomorphic function in a neighborhood of $z_0$ and that $f(z_0)=f'(z_0)=0$ and $f''(z_0)\neq 0$ that is,$f$ has a zero of order $2$ i.e. $f(z)=(z-z_0)^2.G(z)$ where $G(z)$ is a holomorphic function which does not vanish at $z_0$.Now I know that $f$ will locally look like the square function $z\mapsto z^2$ ,which is up to some change of coordinates/local charts.But I am unable to find a suitable change of coordinates map that will take $z_0$ to the point $0$ and $F(z)$ under the change of coordinates will look like $z\mapsto z^2$.
Note:
Here the power $2$ is nothing special.In general if $f$ has a zero of order $k$ then locally $f$ looks like $z\mapsto z^k$ and this is true for any $k\in \mathbb N$,the set of positive integers.
It would be very nice if someone helps me by providing some intuition about how to find that map out?
We have $ G(z_0) = \frac{1}{2}f''(z_0)\neq 0$ by assumption. Let $U$ be a simply connected open neighbourhood of $G(z_0)$, and let $\log \colon U \to \Bbb C$ be a holomorphic determination of the logarithm. On a small enough disk $D(z_0,\varepsilon)$, $G$ takes values in $U$. On this disk, write $f(z) = \varphi(z)^2$ with $\varphi(z) = (z-z_0)\exp\left(\frac{1}{2}\ln G(z)\right)$. Then $\varphi$ is holomorphic on $D(z_0,\varepsilon)$, and straightforward computations show that $(\varphi'(z_0))^2 = G(z_0) \neq 0$. Therefore, $\varphi'(z_0)\neq 0$. By the holomorphic inverse function Theorem, there is $r\in(0,\varepsilon)$ such that $\varphi \colon D(z_0,r) \to V=\varphi(D(z_0,r))$ is a biholomorphism. Note that $0 = \varphi(z_0) \in V$. Hence, $f\circ \varphi^{-1} \colon V \to \Bbb C$ is defined on a neighbourhood of $0$ and has the expression $f(\varphi^{-1}(w)) = \left(\varphi\left(\varphi^{-1}(w)\right)\right)^2 = \omega^2$.
The exact same method works for any $k$.