Show that analytic function $f$ cannot be conformal when $f'(z) = 0$ I am reading over Rudin's discussion of conformal mappings in "Real and Complex Analysis." Rudin states that "no analytic function preserves angles at any point where its derivative is zero. We omit the easy proof of this." So I am trying to fill in the proof.
Suppose $\Omega \subseteq \mathbb{C}$ is open and connnected, and that we are given $z_0 \in \mathbb{C}$ and $f: \Omega \to \mathbb{C}$. Futhermore, suppose that there is some punctured disk $D(z_0, r) \setminus \{z_0\} \subseteq \Omega$ on which $f(z) \neq f(z_0)$. Then Rudin defines $f$ to be angle preserving at $z_0$ iff
$$\lim_{r \to 0} e^{-i\theta}\frac{f(z_0 + re^{i\theta}) - f(z_0)}{|f(z_0 + re^{i\theta}) - f(z_0)|}$$
exists and is independent of $\theta$. 
So, with the additional assumption that $f'(z_0) = 0$, I want to suppose that the limit above exists (and is independent of $\theta$), and then arrive at some contradiction. 
Hints or solutions are greatly appreciated.
 A: Trying to derive a contradiction should be the last resort. Instead we should use what we know about analytical functions.
We may assume $z_0=0$ and $f(0)=f'(0)=0$. If $f$ is not $\equiv0$ (a tacit assumption here) then there is an $n\in{\mathbb N}_{\geq2}$ and an analytic function $g$ with $g(0)=ae^{i\alpha}$, $\ a>0$, such that
$$f(z)=z^n g(z)$$
in a neighborhood of $0$. It follows that
$$e^{-i\theta}\ {f(r e^{i\theta})-f(0)\over|f(r e^{i\theta})-f(0)|}=e^{i(n-1)\theta}{g\bigl(re^{i\theta}\bigr)\over\bigl|g\bigl(re^{i\theta}\bigr)\bigr|}=e^{i(n-1)\theta}\bigl(e^{i\alpha}+o(1)\bigr)\qquad(r\to0)\ .$$
Since $n\geq2$ the  limit   in question depends definitely on $\theta$. 
We had to use extra knowledge about analytic functions. The real $C^1$ map  $${\bf f}(x,y):=(x^2+y^2)(x,y)$$ has differential  $d{\bf f}(0,0)=0$ and preserves angles there.
A: Set $g(z) = f(z_0 + z) - f(z_0)$. Observe that $g$ is analytic on $D(0, r)$, not equal to zero on $D(0, r) \setminus \{0\}$, with $g(0) = g'(0) = 0$. Therefore $g$ has a power series expansion on $D(0,r)$ of the form $\sum_{n = 2}^\infty a_n z^n$. Let $M \ge 2$ be the smallest positive integer such that $a_M \neq 0$ (there must be such an $M$ because $g$ is not constant on $D(0,r)$). For any $\theta$, we have:
$$\lim_{r \to 0}e^{-i\theta}\frac{f(z_0 + re^{i\theta}) - f(z_0)}{|f(z_0 + re^{i\theta}) - f(z_0)|} = \lim_{r \to 0}e^{-i\theta}\frac{g(re^{i\theta})}{|g(re^{i\theta})|}= \lim_{r \to 0}\frac{(re^{i\theta})^M}{e^{i\theta}|re^{i\theta}|^M}\frac{a_M + a_{M+1}re^{i\theta} + \cdots }{|a_M + a_{M+1}re^{i\theta} + \cdots|} = \frac{a_M}{|a_M|}e^{i(M-1)\theta}.$$
So the limit is not independent of $\theta$, and here is a contradiction.
The key to the argument is that $f'(z_0) = 0$ forces $M \ge 2$, and thus the final limit depends on $\theta$.   
