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My book has given me the definition of the derivative for complex functions in several ways but points to one in particular for its geometric aid, I quote:

In this formulation $f:A\rightarrow\mathbf{C}$ is pronounced differentiable at $z_0$ if $z_0$ is interior to $A$ and if there exists a complex number $c$ with the property that $f(z)=f(z_0)+c(z-z_0)+E(z)$ for every $z$ in $A$, where $E:A\rightarrow\mathbf{C}$ is a function satisfying the condition $\lim\limits_{z\rightarrow{z_0}}\frac{|E(z)|}{|z-z_0|}=0$.

Now, my book claims that this is useful, I quote again:

We might, therefore, reasonably expect the geometric behavior of the mapping $w=f(z)$ for $z$ close to $z_0$ to mimic the behavior of the transformation $w=L(z)=f(z_0)+f'(z_0)(z-z_0)$. The geometry fo $L$ is readily understood: $L$ is just a similarity transformation that rotates the complex plane about the point $z_0$ through the angle $\theta=\arg(f'(z_0))$, then dilates the plane so that each ray emanating from $z_0$ is mapped to itself and all distances get scaled by the factor $|f'(z_0)|$, and finally translates the plane so as to move $z_0$ to $f(z_0)$.

Where I mean the principal argument.

Now I'm not sure how the author sees that $L$ is a similarity transformation that does all that he says it does. I understand why the transformation is as it is, $w=L(z)$, but I don't see how the rest follows. I'd really appreciate some elucidation!

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Multiplication by a complex number is a similarity around the origin; this is a consequence of the polar form of complex numbers.

$L(z)$ is just that pre- and post-composed with translations.

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  • $\begingroup$ Ah, so what you are saying is, $|f'(z_0)| e^{i\cdot\arg{f'(z_0)}}\cdot(z-z_0)$. That makes sense, but what I still don't understand (perhaps more of a vocabulary issue) is what is meant when the author says that the transformation "dilates the plane so that each ray emanating from z0 is mapped to itself". $\endgroup$ – Nobody May 16 '14 at 2:52
  • $\begingroup$ @Nihil: What happens to a ray in the complex plane under a map of the form $z \to rz$ where $r$ is a positive real number? $\endgroup$ – Lee Mosher May 16 '14 at 13:26
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Think of it like this $L(z) = re^{i\theta} (z- z_0) + f(z_0)$.

Start with any arbitrary $z$. First step is translation by $-z_0$. Then the result is rotated by $\theta$ and dilated by $r$. Final step is translation by $f(z_0)$.

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