Nonconformal Mapping

Let $\gamma$ be a path in $\Bbb{C}$, $w = f(z)$, and $\theta$ and $\phi$ be the tangents to the curves $\gamma$ and $f(\gamma)$ at $z_0$ and $w_0$. If there is an $N$ such that

$$f^{(N)}(z_0) \not= 0$$ (1)

$$f^{(n)}(z_0) = 0$$ (2)

for all $n < N$ (or, equivalently, if $f'(z)$ has a zero of order $N-1$), then

$$f(z) = f(z_0) + { f^{(N)}(z_0)\over N!} (z-z_0)^N+ { f^{(N+1)}(z_0)\over (N+1)!} (z-z_0)^{N+1} + \ldots$$ (3)

$$f(z)-f(z_0) = (z-z_0)^N\left[{{f(N)(z_0)\over N!} + {f^{(N+1)}(z_0)\over (N+1)!} (z-z_0)+ \ldots}\right],$$ (4)

so the Argument is

$$\arg[f(z)-f(z_0)] = N \arg(z-z_0) +\arg\left[{{ f(N)(z_0)\over N!}+{ f^{(N+1)}(z_0)\over (N+1)!}(z-z_0)+\ldots}\right].$$ (5)

As $z\to z_0$, $\arg(z-z_0)\to \theta$ and $\vert\arg[f(z)-f(z_0)]\vert \to \phi$,

$$\phi = N\theta + \arg \left[{ f(N)(z_0)\over N!}\right]= N\theta +\arg[f(N)(z_0)].$$ (6)

See also Conformal Transformation