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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

$\displaystyle f^{(N)}(z_0)$ $\textstyle \not=$ $\displaystyle 0$ (1)
$\displaystyle f^{(n)}(z_0)$ $\textstyle =$ $\displaystyle 0$ (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
\end{displaymath} (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],
\end{displaymath} (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].
\end{displaymath} (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)].
\end{displaymath} (6)

See also Conformal Transformation

© 1996-9 Eric W. Weisstein