# Understanding a Similarity Transformation with a Fixed Point from the Definition of the Derivative for Complex Functions

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!

$L(z)$ is just that pre- and post-composed with translations.
• 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". – Nobody May 16 '14 at 2:52
• @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? – Lee Mosher May 16 '14 at 13:26
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 $f(z_0)$.