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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.

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    $\begingroup$ Let $Z_0=0,Z_1=1,Z_2=i$. Then we have a right angle at $Z_0$. The function mapping $z\mapsto z^2$ is holomorphic. However applying it to $Z_0,Z_1,Z_2$ gives the points $0,1,-1$ and our right angle is mapped to a pair of lines meeting at $\pi$ radians. Making $Z_1$, $Z_2$ smaller by any positive real factor will not alter this. $\endgroup$
    – tkf
    Commented Feb 17, 2022 at 4:31
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    $\begingroup$ @tkf I should add $f'(Z_0) \neq 0$. That will fix it right? $\endgroup$
    – zjk
    Commented Feb 17, 2022 at 4:56
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    $\begingroup$ No: $1,2,1+i\mapsto1,4,2i$ under squaring. $\endgroup$
    – tkf
    Commented Feb 17, 2022 at 4:59
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    $\begingroup$ @tkf Shrink by a factor of 0.1. That'll be $(1, 1.1, 1 + 0.1i) \mapsto (1, 1.21, 0.99 + 0.2i)$ seems kind of similar $\endgroup$
    – zjk
    Commented Feb 17, 2022 at 5:13
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    $\begingroup$ @zjk I took "sort of claim" to mean you thought the result would hold exactly, subject to some hypothesis'. Perhaps "claim sort of" would have conveyed your intuition better. In any case it has been made precise in the excellent answer provided (+1). $\endgroup$
    – tkf
    Commented Feb 17, 2022 at 13:41

1 Answer 1

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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 $\lim_{z\to c}g(z)=0$.

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  • $\begingroup$ Since $g(c)$ is technically undefined, I might modify the last part to just say that $\lim_{z\to c}g(z)=0$. So as $y$ and $z$ approach $c$ (regardless of the relative rates of approach), the ratio in question approaches 1. $\endgroup$
    – Karl
    Commented Feb 17, 2022 at 6:34
  • $\begingroup$ Yes you're right $\endgroup$ Commented Feb 17, 2022 at 6:37
  • $\begingroup$ I think your answer is more like a proof of why the claim is valid (which is helpful). I'm looking for something like a modified definition of similar triangles. $\endgroup$
    – zjk
    Commented Feb 17, 2022 at 6:51
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    $\begingroup$ $\dfrac{f'(c)+g(z)}{f'(c)+g(y)}$ can be viewed as measuring "how similar" the triangles are. If the fraction is closer to $1$, the triangles are "more similar". $\endgroup$ Commented Feb 17, 2022 at 7:03

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