complex differentiability of conformal function

Question

I have studied a complex analysis with textbook "Function Theory of One Complex Variable" by Robert Everist Greene, Steven George Krantz. While studying about "Conformality" and several exercises regarding it, I got a question. The question is that : Assume that $F : U \rightarrow \mathbb{C}$ is $C^1$ on an open set $U$. Let $P$ be in $U$. Assume that $F$ "stretches equally in all directions at $P$". (The size of all directional derivatives is the same for all unit modulus.) Prove that either $F'(P)$ exists or $\bar{F'}(P)$ exists

My attempts to solve it are followings : Let $u$ and $v$ are a real and imaginery part functions of $F$, repectively. Then I only need to show that these function satify the Cauchy-Riemann Equations. Then it induces that the comlex valued $C^{1}$ function $F$ satisfies Cauchy Riemann equations, so it is holomorphic at $P$ and differentiable at $P$ obviously.

But, with the given informations, I dont know how to induce the Cauchy Riemann Equations. I need some proper advice for my studies.

Answer 1

We use the following easy fact: assume $w,z$ complex numbers s.t. $|z\cos \theta + w\sin \theta |$ doesn't depend on $\theta$, then $w= \pm iz$

(proof: wlog assume $w,z \ne 0$ as otherwise statement trivial and applying this with $\theta=0, \frac{\pi}{2}$ we get $|w|=|z|$ so $w=\alpha z, |\alpha|=1$; but then squaring, $| \cos \theta+ \alpha\sin \theta |^2=1+2(\sin \theta \cos \theta)\Re \alpha$ obviously doesn't depend on $\theta$ iff $\Re \alpha =0$, hence $\alpha =\pm i$ as required)

Applying this with $z=f_x(P), w=f_y(P)$ we get from the hypothesis and the fact above that $f_y(P) =\pm i f_x(P)$. This clearly implies $f$ analytic at $P$ if we have the plus sign and $f$ conjugate analytic if the sign is minus, so we are done !

Answer 2

Three years late, but someone may find this solution useful too. Let $A=Df(p)$ denote the 2x2 matrix that represents the derivative (calculus sense) of $f$ at $p$. The assumption on modulus of directional derivatives is just saying that for all $v$, we have $|Av|= c$ for a constant $c$. Case $c=0$ is easy, so assume otherwise.

Fact 1 (Linear algebra, in all dimensions): The only $n\times n$ matrices for which $|Av|$ is independent of $v$ are the ones that are scalar multiples of orthogonal matrices. So, $A=c' O$, where $OO^t=1$, the identity matrix.

Fact 2: Suppose $f$ is differentiable at $p$. Then the following are equivalent: 1) $f$ is complex differentiable at $p$; 2) its partial derivatives satisfy Cauchy-Riemann equations at $p$; 3) The derivative $Df(p)$ is a multiple of an orthogonal matrix and has positive determinant.

Equivalence between 2 and 3 is seen by noticing that in any orthogonal matrix in 2-D, if the first column is $(a,b)$ then the second must be either $(b,-a)$ or $(-b,a)$, which is decided by the sign of the determinant.

Now, the claim in the post follows by these facts and that if $Df$ has negative determinant, then $D\overline{f}(p)$ would "fix" it.