Similarly to this problem:If there exists $\lim_{\Delta z \to 0} |\frac{\Delta w}{\Delta z}|$, then $w$ or $\bar{w}$ is holomorphic., this is problem no. 113 from "A Collection of Problems on Complex Analysis" by L.I. Volkovyskii, G.L. Lunts and I.G. Aramanovich. Here's the full formulation:
At the point $z_{0}$ the function $w=f(z)=u(z)+iv(z)$ possesses the following properties:
$u$ and $v$ are ($\mathbb{R}$-)differentiable
the limit $\lim\limits_{\Delta z \to 0} \arg\frac{\Delta w}{\Delta z}$ exists.
Prove that $f$ is ($\mathbb{C}$-)differntiable at $z_{0}$.
Again, $\Delta w = f(z_{0} +\Delta z) - f(z_{0})$.
After solving the linked problem, I thought I'd have a field day with this one. However, I know even less about this problem than the previous one (even with the previous one solved), because I don't know much about the funciton $\arg$ (as in, I don't know a nice way to manipulate it algebraically). Denote the limit in 2. as $A$.
My attempt (again) was to try plugging in certain directions: for $\Delta z = \Delta x$, I have $A = \arg (u_{x} - iv_{x}$), for $\Delta z = i\Delta y$, I have $A = \arg(v_{y}+iu_{y})$. For $\Delta z = \Delta x + i \Delta x$, I get $A = \arg (u_{x}+u_{y}+v_{x}+v_{y}+i(u_{x}+u_{y}-v_{x}-v_{y}))$. Then, for $\Delta z = \Delta x - i \Delta x$, I get $A=\arg(u_{x}+u_{y}-v_{x}-v_{y}-i(u_{x}+u_{y}+v_{x}+v_{y}))$.
The only thing that occurs to me from here is to conclude, for example, that $\arg(\frac{u_{x}-iv_{x}}{v_{y}+iu_{y}})=0,$ so $\frac{u_{x}-iv_{x}}{v_{y}+iu_{y}}=c \in \mathbb{R}$, so then $u_{x} = cv_{y}$, $-v_{x} = cu_{y}$, but I don't know how to prove that $c=1$. Plugging in the last two identities for $A$ doesn't lead me anywhere.
Hopefully this won't be marked as a duplicate, even though the idea is probably similar.
The real differentiability of $f$ at $z_0$ implies there exist $A,B\in \mathbb C$ such that
$$f(z_0+h)-f(z_0) = Ah +B\bar h + o(z)\,\, \text { as } h\to 0.$$
This implies
$$\tag 1 \frac{f(z_0+h)-f(z_0)}{h}= A +B\frac{\bar h}{h} + o(1)\,\, \text { as } h\to 0.$$
For any fixed $t\in \mathbb R,$ let $h=re^{it}$ in $(1),$ and let $r\to 0^+.$ We get the limit
$$\tag 2 A +Be^{-2it}.$$
Now by hypothesis, the argument of $(1)$ has a limit as $h\to 0.$ This implies all the numbers in $(2),$ for $t\in \mathbb R,$ lie on the same ray from $0.$ But if $B\ne 0,$ then the numbers in $(2)$ form a circle of radius $|B|>0$ and center $A.$ Clearly no circle of positive radius can lie within a ray. Thus $B=0.$ Hence the limit in $(1)$ as $h\to 0$ equals $A,$ which implies $f'(z_0) = A$ as desired.