- Let \(f\) be a complex-valued \(C ^ { 1 }\) function defined in the neighborhood of a point \(z _ { \mathrm { 0 } }\) . There are several notions closely related to conformality at \(z _ { 0 }\) . We say that \(f\) is isogonal at \(z _ { 0 }\) if whenever \(\gamma ( t )\) and \(\eta ( t )\) are two smooth curves with \(\gamma ( 0 ) =\) \(\eta ( 0 ) = z _ { 0 }\) , that make an angle θ there \(( | \theta | < \pi )\) , then \(f ( \gamma ( t ) )\) and \(f ( \eta ( t ) )\) make an angle of \(\theta ^ { \prime }\) at \(t = 0\) with \(| \theta ^ { \prime } | = | \theta |\) for all θ. Also, \(f\) is said to be isotropic if it magnifies lengths by some factor for all directions emanating from \(z _ { 0 }\) , that is, if the limit
\[ \lim _ {r \to 0} \frac {| f (z _ {0} + r e ^ {i \theta}) - f (z _ {0}) |}{r} \]
exists, is non-zero, and independent of \(\theta .\)
Then \(f\) is isogonal at \(z _ { 0 }\) if and only if it is isotropic at \(z _ { \mathrm { 0 } } ;\) moreover, \(f\) is isogonal at \(z _ { 0 }\) if and only if either \(f ^ { \prime } ( z _ { 0 } )\) exists and is non-zero, or the same holds for \(f\) replaced by \({ \overline { { f } } } .\)
Solution
Let \(A=Df(z_0):\mathbb R^2\to\mathbb R^2\) be the real differential. For a unit vector \(v=e^{i\theta}\), \[ f(z_0+rv)-f(z_0)=rAv+o(r), \] so \[ \lim_{r\to0}\frac{|f(z_0+rv)-f(z_0)|}{r}=|Av|. \] Thus \(f\) is isotropic at \(z_0\) exactly when there is a number \(\lambda>0\) such that \[ |Av|=\lambda|v|\qquad(v\in\mathbb R^2). \tag{1} \] By polarization, (1) is equivalent to \[ \langle Av,Aw\rangle=\lambda^2\langle v,w\rangle \] for all \(v,w\), or equivalently \[ A^TA=\lambda^2I. \tag{2} \] Hence \(A=\lambda Q\) with \(Q\in O(2)\). Such a map preserves the absolute value of every angle, so isotropy implies isogonality.
Conversely, suppose \(f\) is isogonal. Then \(A\) cannot kill a nonzero direction, since otherwise the image tangent direction of a curve with that tangent would be undefined. Choose orthonormal vectors \(e_1,e_2\). Isogonality gives \(Ae_1\perp Ae_2\). Applying it also to the pair \(e_1+e_2,e_1-e_2\), which are perpendicular, yields \[ 0=\langle A(e_1+e_2),A(e_1-e_2)\rangle =|Ae_1|^2-|Ae_2|^2. \] Thus \(|Ae_1|=|Ae_2|=\lambda>0\), and since their images are orthogonal, (2) follows. Therefore \(f\) is isotropic.
It remains to identify the two possibilities for \(A\). Every \(Q\in O(2)\) has determinant \(1\) or \(-1\). If \(\det Q=1\), then \(Q\) is rotation by some angle and, after identifying \(\mathbb R^2\) with \(\mathbb C\), \[ Av=cv \] for some \(c\in\mathbb C\setminus\{0\}\). Hence the complex derivative \(f'(z_0)\) exists and equals \(c\).
If \(\det Q=-1\), then \(Q\) is a rotation followed by reflection, so \[ Av=c\overline v \] for some \(c\ne0\). Equivalently, \[ D\overline f(z_0)(v)=\overline c\,v, \] so \(\overline f\) has a nonzero complex derivative at \(z_0\).
Conversely, either of these two differential forms is a nonzero similarity and therefore preserves absolute angles and magnifies every direction by the same factor. Thus the three stated conditions are equivalent.