Let $f:\Omega \to \mathbb C$, where $\Omega$ is a region in $\mathbb C$, and $z_0 \in \Omega$.

Suppose that $\displaystyle \lim_{z \to z_0} \frac {|f(z)-f(z_0)|}{|z-z_0|}=k$ for some constant k regardless of the direction from which $z$ is approaching to $z_0$.

Does it mean that $f(z)$ is analytic?

I don't think this is true because $\displaystyle \lim_{z \to z_0} \frac {|f(z)-f(z_0)|}{|z-z_0|}=\displaystyle \lim_{z \to z_0} \frac {|\bar f(z)-\bar f(z_0)|}{|z-z_0|}$ and $f, \bar f$ cannot be analytic simultaneously unless they are constants. So this basically would basically mean that all analytic functions are constants, which is clearly wrong.

However, Ahlfors writes (At least, I interpreted so) in his textbook Complex Analysis that this property implies the differentiability of $f$ with "additional regularity assumptions". (P.73-74 3rd edition)

Conversely, it is clear that both kinds of conformality (preserving the angle and the size) together imply the existence of $f'(z_o)$. It is less obvious that each kind will separately imply the same result, at least under additional regularity assumptions

I guess I interpreted wrong way but I don't know what's wrong with my argument. Could anyone help me with this?

  • $\begingroup$ You did not prove that the angle is preserved. $\endgroup$
    – akkkk
    Dec 3 '12 at 15:59
  • $\begingroup$ @akkkk He says "each kind will separately imply the same result". So I thought if the size is preserved, it implies the same result $\endgroup$
    – Tengu
    Dec 3 '12 at 16:05
  • 1
    $\begingroup$ "under additional regularity assumptions". $\endgroup$
    – akkkk
    Dec 3 '12 at 16:06
  • $\begingroup$ @akkkk What's regularity here? I thought the regularity is a property of an arc that means that it has a continuous derivative that does not vanish. So I guess it has nothing to with angle here? $\endgroup$
    – Tengu
    Dec 3 '12 at 16:10
  • $\begingroup$ Please avoid using \displaystyle in the title. $\endgroup$
    – Asaf Karagila
    Dec 3 '12 at 22:24

You are right that $\bar f$ satisfies the condition $\lim_{z \to z_0} \frac {|f(z)-f(z_0)|}{|z-z_0|}=k$. This is however the only exception as the following theorem shows.

Theorem. Suppose that $f$ satisfies the condition $$\lim_{z \to z_0} \frac {|f(z)-f(z_0)|}{|z-z_0|}=k$$ in some region $\Omega$. Assume additionally that $f$ is sufficiently smooth (e.g. $f \in C^1({\mathbb R}^2, {\mathbb R}^2)$) and $k\neq 0$ (at all points $z_0\in \Omega$). Then either $f$ is holomorphic in $\Omega$ or $\bar f$ is holomorphic in $\Omega$.

(The additional assumption can be relaxed.)

Proof: Write $f(z) = u(z) + i v(z)$ and $z = x+ i y$. Consider the Jacobian matrix $J_f$ of $$J_f = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix}. $$

Note that for every $(a,b) \in {\mathbb R}^2 \setminus \{0\}$, $$k = \lim_{t \to 0} \frac {|f(z_0 + (a+bi)t)-f(z_0)|}{|a+bi|\cdot |t|}=\frac{\left|J_f \begin{pmatrix} a\\ b\end{pmatrix}\right|}{|(a,b)|}.$$ Therefore, for every $v\in {\mathbb R}^2$, $|\left(\frac{1}{k} J_f\right) v| = |v|$. Thus $\frac{1}{k} J_f$ is an orthogonal operator.

There are two possibilities: $\frac{1}{k} J_f$ is either a rotation or reflection matrix. So if $\det(J_f) > 0$ then $J_f$ is a scaled rotation matrix: $$J_f = \begin{pmatrix} \alpha & \beta\\ -\beta & \alpha \end{pmatrix}. $$ Thus $f$ satisfies Cauchy–Riemann equations. Otherwise, $J_f$ is a scaled reflection matrix: $$J_f = \begin{pmatrix} \alpha & \beta\\ \beta & -\alpha \end{pmatrix}, $$ and $\bar f$ satisfies Cauchy–Riemann equations.

Note also that since $k\neq 0$ either $f$ satisfies Cauchy–Riemann equations at all points $z\in \Omega$, or $\bar f$ satisfies Cauchy–Riemann equations at all points $z\in \Omega$. It also follows that $f'(z)\neq 0$.

  • $\begingroup$ Thank you for your explanation. Now I understand that either $f$ or $\bar f$ is holomorphic. Then "It is less obvious that each kind will separately imply the same result, at least under additional regularity assumption" is not really true? $\endgroup$
    – Tengu
    Dec 3 '12 at 23:22
  • $\begingroup$ It depends on what the words “additional regularity assumption” mean :). But strictly speaking you are right this sentence is not true. $\endgroup$
    – Yury
    Dec 3 '12 at 23:25
  • $\begingroup$ Thank you very much! By the way, what does "regularity" here mean? Since it involves arcs, I thought it means that the arcs need to be regular. It's because he divides an equation by the derivative of an arc in his poof. I thought that's why he needed regularity assumption. $\endgroup$
    – Tengu
    Dec 3 '12 at 23:56
  • $\begingroup$ Usually, a regularity condition requires that a certain object is “nice” in a certain way. E.g. in the theorem I proved the regularity condition is that “$f$ is sufficiently smooth and $k\neq 0$” (this condition can be relaxed). The author probably had a similar condition in mind. I don't think it has anything to do with arcs. $\endgroup$
    – Yury
    Dec 4 '12 at 0:10
  • $\begingroup$ Thank you. Hm.. The reason why I thought of an arc is because the author defined a regular arc in a previous section. Anyway I didn't know that "regularity" means such a thing. Thanks again :) $\endgroup$
    – Tengu
    Dec 4 '12 at 0:12

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