The question is :
Prove that a holomorphic function $\varphi:\Omega\to\Omega$ is linear if $\varphi(z_0)=z_0$ and $\varphi'(z_0)=1$ for $z_0\in\Omega$, which is an bounded open subset of $\mathbb C$.
(Stein & Shakarchi, Complex Analysis, Chapter 2, Exercise 9)
I know that this problem and perfect answer are already in MathStackexchange and solved by that direction(defining $\varphi^k$ and applying Cauchy inequalities), and I'm approaching at different point.
First I developed a power series expansion of $\varphi$ near $z_0$ of radius $R$, and the radius must be bounded since $\Omega$ is bounded. Then by the given conditions, the restricted function $\varphi_0$ is like below. $$\varphi_0:D_R(z_0)\subset\Omega\to\Omega,\quad\varphi_0(z)=z_0+(z-z_0)+(z-z_0)^2\sum_{n=0}^{\infty}a_{n+2}(z-z_0)^n$$ So if we define $\sum a_{n+2}(z-z_0)^n=(\varphi_0(z)-z)/(z-z_0)^2=g(z)$, $g(z)$ must be identically zero and then $\varphi_0(z)$ be a linear function. Then we will be able to use analytic continuation to expand this function over $\Omega$.
First I assumed that $g(z)$ is not identically zero so there will be small $r>0$ such that for all $z\in D_r(z_0)\backslash\{z_0\}$, $g(z)\neq0$. I want to show that in this deleted disc, $g(z)$ converges to zero and the assumption is false.
By the way, since $\varphi'(z_0)=1$, I know that the limit below works, $$\lim_{z\to z_0}(z-z_0)g(z)=\lim_{z\to z_0}\left(\dfrac{\varphi_0(z)-\varphi_0(z_0)}{z-z_0}-1\right)=0$$ but the factor $(z-z_0)$ is still in the formula.
How can I prove that $g$ is identically zero in the disc?
Thanks.