# holomorphic mapping of curve and tangent

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 $\gamma_{w}$.

We suppose still that the curve $\gamma$ has a tangent line  at the point $z_{0}$ and that the value of the derivative  $f^{\prime}$ has in $z_{0}$ a nonzero value

 $\displaystyle f^{\prime}(z_{0})\,=\,\varrho e^{i\omega}.$ (1)

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

 $\Delta z\,=\,ke^{i\alpha},\quad\Delta w\,=\,k_{w}e^{i\alpha_{w}},$

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

 $\lim_{\Delta z\to 0}\frac{\Delta w}{\Delta z}\;=\;f^{\prime}(z_{0}).$

This implies, by (1), that

 $\displaystyle\lim_{\Delta z\to 0}\frac{k_{w}}{k}\;=\;\varrho.$ (2)

From this we infer, because  $\varrho\neq 0$  that, up to a multiple of $2\pi$,

 $\displaystyle\lim_{\Delta z\to 0}(\alpha_{w}-\alpha)\;=\;\omega.$ (3)

But the limit of $\alpha$ is the slope angle $\varphi$ of the tangent  of $\gamma$ at $z_{0}$.  Hence (3) implies that

 $\displaystyle\varphi_{w}\;=\;\lim_{\Delta z\to 0}\alpha_{w}\;=\;\varphi+\omega.$ (4)

Accordingly, we have the

Theorem 1.  If a curve $\gamma$ has a tangent line in a point $z_{0}$ where the derivative $f^{\prime}$ 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

 $\omega\;=\;\arg f^{\prime}(z_{0}).$

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  :

 $\displaystyle\lim_{\Delta z\to 0}\frac{s_{w}}{s}\;=\;|f^{\prime}(z_{0})|.$ (5)

Conformality

If we have besides $\gamma$ another curve $\gamma^{\prime}$ 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):

 $\varphi_{w}\;=\;\varphi+\omega,\quad\varphi_{w}^{\prime}\;=\;\varphi^{\prime}+\omega$

By subtracting we obtain

 $\displaystyle\varphi_{w}^{\prime}-\varphi_{w}\;=\;\varphi^{\prime}-\varphi,$ (6)

whence we have the

Theorem 2.  The mapping created by the holomorphic function $f$ preserves the magnitude of the angle between two curves in any point $z$ where  $f^{\prime}(z)\neq 0$.  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 directly conformal.  If the orientation were reversed the mapping were called inversely conformal; in this case $f$ were not holomorphic but

Title holomorphic mapping of curve and tangent HolomorphicMappingOfCurveAndTangent 2013-03-22 18:42:19 2013-03-22 18:42:19 pahio (2872) pahio (2872) 9 pahio (2872) Topic msc 53A30 msc 30E20 directly conformal