Rigorous statement on that two small trianges, one being the image of the other by a holomorphic function, are similar Let $w = f(z)$ be a non-constant holomorphic function. Let $Z_0$ be a point in the complex plane, $Z_1$ are $Z_2$ are points in the neighbourhood of $Z_0$.  Let $W_0, W_1, W_2$ be the corresponding points in the image complex plane of $w = f(z)$.
One can sort of claim that $\triangle W_0 W_1 W_2$ and $\triangle Z_0 Z_1 Z_2$ are similar as long as the neighbourhood is small.
This statement is visually intuitive. I'm wondering how can one make this statement rigorous like the $\epsilon-\delta$ way of defining limit. Ideally the rigorous statement would keep the intuition.
 A: Let's fix a point $c$ and assume, as you say, that $f'(c)\neq0$.
Let's fix $r\in\mathbb{R}^+$ small enough so that $f$ is an injective function on $\{z\in\mathbb{C} :|z-c|<r\}$. This will be our neighbourhood of $c$.
(If you don't understand why this choice of $r$ is always possible, you might want to read the following: If $f:U\to V$ is holomorphic and $f'(z)\neq 0$ for all $z\in U$, then$f$ is locally bijective.)
Let $y$ and $z$ be variables representing points on this neighbourhood. Also assume that $y\neq c$ and $z\neq c$, otherwise the three points won't form a triangle. The injectivity of $f$ is useful because $f(y)\neq f(c)$ and $f(z)\neq f(c)$.
Notice that $\dfrac{z-c}{y-c}$ gives you all the information you need to know about the triangle $CYZ$, if similar triangles are considered the same. The fraction shows you not only the angle, but also the ratio between the lengths of $CZ$ and $CY$.
Now you just compare that with $\dfrac{f(z)-f(c)}{f(y)-f(c)}$.
We have:
$\left.\dfrac{f(z)-f(c)}{f(y)-f(c)} \middle/ \dfrac{z-c}{y-c} \right.=
\left.\dfrac{f(z)-f(c)}{z-c} \middle/ \dfrac{f(y)-f(c)}{y-c} \right.=\dfrac{f'(c)+g(z)}{f'(c)+g(y)}$
where $g(z)=\dfrac{f(z)-f(c)}{z-c}-f'(c)$.
Notice that $\lim_{z\to c}g(z)=0$.
