holomorphic mapping of curve and tangent
HolomorphicMappingOfCurveAndTangent
2009-01-06 14:22:29
2009-01-07 13:44:36
Topic
conformal mapping
directly conformal
conformal mapping
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\newtheorem*{thmplain}{Theorem}
Let $D$ be a domain of the complex plane and the function \,$f\!:\,D \to \mathbb{C}$\, be holomorphic.\, Then for each point $z$ of
$D$ there is a corresponding point \,$w = f(z)\,\in \mathbb{C}$;\, we think that $z$ and $w$ both lie in their own complex planes, $z$-plane and $w$-plane.
Since $f$ is continuous in $D$, if $z$ draws a continuous curve $\gamma$ in $D$ then its image point $w$ also draws a continuous curve $\gamma_w$.\, Let $z_0$ and $z_0\!+\!\Delta z$ be two points on $\gamma$ and $w_0$ and
$w_0\!+\!\Delta w$ their image points on $\gamma_w$.
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We suppose still that the curve $\gamma$ has a tangent line at the point $z_0$ and that the value of the derivative $f'$ has in $z_0$ a nonzero value
\begin{align}
f'(z_0) \,=\, \varrho e^{i\omega}.
\end{align}
If the slope angles of the secant lines \,$(z_0,\,z_0\!+\!\Delta z)$\, and\, $(w_0,\,w_0\!+\!\Delta w)$\, are $\alpha$ and $\alpha_w$, then we have
$$\Delta z \,=\, ke^{i\alpha}, \quad \Delta w \,=\, k_we^{i\alpha_w},$$
and the difference quotient of $f$ has the form
$$\frac{\Delta w}{\Delta z} \;=\;
\frac{f(z_0\!+\!\Delta z)-f(z_0)}{\Delta z} \,=\, \frac{k_w}{k}e^{i(\alpha_w-\alpha)}.$$
Let now\, $\Delta z \to 0$.\, Then the point $z_0\!+\!\Delta z$ tends on the curve $\gamma$ to $z_0$ and
$$\lim_{\Delta z \to 0}\frac{\Delta w}{\Delta z} \;=\; f'(z_0).$$
This implies, by (1), that
\begin{align}
\lim_{\Delta z \to 0}\frac{k_w}{k} \;=\; \varrho.
\end{align}
From this we infer, because\, $\varrho \neq 0$\, that, up to a multiple of $2\pi$,
\begin{align}
\lim_{\Delta z \to 0}(\alpha_w-\alpha) \;=\; \omega.
\end{align}
But the limit of $\alpha$ is the slope angle $\varphi$ of the tangent of $\gamma$ at $z_0$.\, Hence (3) implies that
\begin{align}
\varphi_w \;=\; \lim_{\Delta z \to 0}\alpha_w \;=\; \varphi+\omega.
\end{align}
Accordingly, we have the
\textbf{Theorem 1.}\, If a curve $\gamma$ has a tangent line in a point $z_0$ where the derivative $f'$ does not vanish, then the image curve $f(\gamma)$ also has in the corresponding point $w_0$ a certain tangent line with a direction obtained by rotating the tangent of $\gamma$ by the angle
$$\omega \;=\; \arg f'(z_0).$$\\
If the curve $\gamma$ is smooth, then also $\gamma_w$ is smooth, and it follows easily from (2) the corresponding limit equation between the arc lengths:
\begin{align}
\lim_{\Delta z \to 0}\frac{s_w}{s} \;=\; |f'(z_0)|.
\end{align}
\textbf{Conformality}
If we have besides $\gamma$ another curve $\gamma'$ emanating from $z_0$ with its tangent, the mapping $f$ from $D$ in $z$-plane to $w$-plane gives two curves and their tangents emanating from $w_0$.\, Thus we have two equations (4):
$$\varphi_w \;=\; \varphi+\omega, \quad \varphi_w' \;=\; \varphi'+\omega$$
By subtracting we obtain
\begin{align}
\varphi_w'-\varphi_w \;=\; \varphi'-\varphi,
\end{align}
whence we have the
\textbf{Theorem 2.}\, The mapping created by the holomorphic function $f$ preserves the magnitude of the angle between two curves in any point $z$ where\, $f'(z) \neq 0$.\, The equation (6) tells also that the orientation of the angle is preserved.\\
The facts in Theorem 2 are expressed so that the mapping is {\em directly conformal}.\, If the orientation were reversed the mapping were called {\em inversely conformal}; in this case $f$ were not holomorphic but {\em antiholomorphic.}