If there exists $\lim\limits_{\Delta z \to 0} |\frac{\Delta w}{\Delta z}|$, then $w$ or $\bar{w}$ is holomorphic. This is problem no. 112 from "A Collection of Problems on Complex Analysis" by L.I. Volkovyskii, G.L. Lunts and I.G. Aramanovich.

At the point $z_{0}$ the function $w=f(z)$ possesses the following properties:

*

*$u$ and $v$ are ($\mathbb{R}$-)differentiable


*the limit $\lim\limits_{\Delta z \to 0} |\frac{\Delta w}{\Delta z}|$ exists.
Prove that either $f(z)$ or $\bar{f}(z)$ is differentiable (complex-differentiable) at $z_{0}$.

The convention here is that $u = \Re f$, $v=\Im f$, and $\Delta z = \Delta x + i \Delta y = z-z_{0}$, and $\Delta w = w-w_{0} = f(z_{0}+\Delta z) - f(z_{0}) = f(z)-f(z_{0})$.
So $f=u+iv$, $\bar{f} = u -iv$, etc.
Here's what I've tried: if the limit from 2. is $A$, for $\Delta z = \Delta x$, we have $A^2 = u_{x}^2+v_{x}^2$, and if $\Delta z = i \Delta y$, we have $A^2 = v_{y}^2 + u_{y}^2$. This follows from the fact that $$|\frac{\Delta w}{\Delta z}| = \frac{(\Delta u \Delta x+ \Delta v \Delta y) +i(\Delta u \Delta y - \Delta v \Delta x)}{|\Delta z|^2}.$$
Are there any other directions I should try except $(1,0)$ and $(0,1)$? Would it even yield any different results?
Obviously, $f$ is defined in an open ball around $z_{0}$ of positive radius.
 A: I solved the problem after (many) tips from Somos (so a big thank you to Somos).
By setting a fixed (arbitrary) direction $\Delta z = h(\alpha, \beta)$, $h \in \mathbb{R}$, we get $\Delta x = h \alpha$, $\Delta y = h \beta$, $\Delta u = u_{x}h\alpha + u_{y}h\beta$, and the same for $\Delta v$.
Thus, the equality $A^2 = \lim_{\Delta z \to 0} |\frac{\Delta w}{\Delta z}|^2$ becomes (after a lot of addition, subtraction, multiplication and division (this can probably be instantly seen from the fact that the partial derivatives form a base of the tangent space at a point, but I didn't notice that on time), and by setting $\alpha^2 + \beta^2 = 1$, which we can do WLOG since $h$ varies)
$$ A^2 = (\alpha u_{x} + \beta u_{y})^2 + (\alpha v_{x} + \beta v_{y})^2.$$
Now, if we set $\alpha = 1$, $\beta = 0$, we get $u_{x}^2 + v_{x}^2 = A^2$, and if we set $\alpha = 0$, $\beta = 1$, we get $u_{y}^2 + v_{y}^2 = A^2$. So we have $u_{x}^2+v_{x}^2 = u_{y}^2+v_{y}^2$ , and by setting $\alpha = \beta = \frac{1}{\sqrt{2}}$, we get $u_{x}u_{y}+v_{x}v_{y} = 0$. We can assume, for example, that $u_{y} \neq 0$ (if $u_{x} = u_{y} = v_{x} = v_{y} = 0$, then $f = const$, so $f$ is holomorphic then), and express $u_{x}$ as $-\frac{v_{x}v_{y}}{u_{y}}$, and by plugging it into the first equation, we get $v_{x}^2 = u_{y}^2$. Then we have $v_{x} = \pm u_{y}$, and from there we get $v_{y} = \mp u_{x}$, so either $f$ is differentiable or $\bar{f}$ is differentiable at $z_{0}$.
A: Here's another approach: Suppose WLOG $z_0=0$ and $f(0)=0.$ The real differentiability of $f$ at $0$ implies there exist $A,B\in \mathbb C$ such that
$$f(z) = Az +B\bar z + o(z)\,\, \text { as } z\to 0.$$
This implies $\dfrac{f(z)}{z}= A +B\dfrac{\bar z}{z} + o(1).$ Letting $z=re^{it}$ and taking absolute values then gives
$$\left | \frac{f(re^{it})}{re^{it}}\right | = |A + Be^{-2it} +o(1)|.$$
Now letting $re^{it}\to 0,$ we see that $|A + Be^{-2it}|$ is independent of $t.$ Verify that this is impossible if both $A,B\ne 0.$ Thus either $A = 0,$ in which case $(\bar f)'(0) = \bar B,$ or $B = 0,$ in which case $f'(0) = A.$  
