Proving a sufficient condition for complex differentiability I'm trying to show that given $f=u+iv:\mathbb{C}\to\mathbb{C}$
  and $z_{0}\in\mathbb{C}$
  if $u,v$
  are differentiable (as functions $\mathbb{R}^{2}\to\mathbb{R})$
  at $z_{0}=\left(x_{0},y_{0}\right)$
  and the limit $$\lim\limits _{z\to z_{0}}\mbox{arg}\left(\frac{f(z)-f\left(z_{0}\right)}{z-z_{0}}\right)$$
  exists then $f$
  is complex differentiable at $z_{0}$
 . I've reached a point where it would suffice if I could show that under these conditions I can conclude that the limit $$\lim\limits _{z\to z_{0}}\left|\frac{f\left(z\right)-f\left(z_{0}\right)}{z-z_{0}}\right|$$
  exists. Now we can see that: $$\lim\limits _{z\to z_{0}}\left|\frac{f\left(z\right)-f\left(z_{0}\right)}{z-z_{0}}\right|=\lim_{z\to z_{0}}\sqrt{\frac{\left(u\left(z\right)-u\left(z_{0}\right)\right)^{2}+\left(v\left(z\right)-v\left(z_{0}\right)\right)^{2}}{\left|z-z_{0}\right|^{2}}}=\sqrt{\lim_{z\to z_{0}}\frac{\left(u\left(z\right)-u\left(z_{0}\right)\right)^{2}+\left(v\left(z\right)-v\left(z_{0}\right)\right)^{2}}{\left|z-z_{0}\right|^{2}}}$$
 So it would suffice if I could show that these following limits exist: $$\lim_{z\to z_{0}}\frac{\left(u\left(z\right)-u\left(z_{0}\right)\right)^{2}}{\left|z-z_{0}\right|^{2}}\quad,\quad\lim_{z\to z_{0}}\frac{\left(v\left(z\right)-v\left(z_{0}\right)\right)^{2}}{\left|z-z_{0}\right|^{2}}$$
 I've tried getting to the existence of these limit from the differentiability of u,v
  but without success.
Note: when wirting $u\left(z\right),v\left(z\right)$
  the intention is of course $ u\left(x,y\right),v\left(x,y\right)
 $
  as these are not complex function but functions $\mathbb{R}^{2}\to\mathbb{R}$
 . It's also worth noting that the norm on $\mathbb{R}^{2}$
  is exactly the complex modulus so one can rewrite that we need the existence of the following limits: $$\lim_{\left(x,y\right)\to\left(x_{0},y_{0}\right)}\frac{\left(u\left(x,y\right)-u\left(x_{0},y_{0}\right)\right)^{2}}{\left\Vert \left(x,y\right)-\left(x_{0},y_{0}\right)\right\Vert ^{2}}\quad,\quad\lim_{\left(x,y\right)\to\left(x_{0},y_{0}\right)}\frac{\left(v\left(x,y\right)-v\left(x_{0},y_{0}\right)\right)^{2}}{\left\Vert \left(x,y\right)-\left(x_{0},y_{0}\right)\right\Vert ^{2}}$$
Help would be appreciated!
 A: As the concepts we are talking about here are translation invariant we may assume $z_0=f(z_0)=0\in{\mathbb C}$. The assumption that $f=u+iv$ is real differentiable at $(0,0)=0\in{\mathbb C}$ implies
$$u(x,y)=u_1 x+u_2 y+o\bigl(|z|\bigr),\quad v(x,y)=v_1 x+v_2 y+o\bigl(|z|\bigr)\qquad (z\to0)$$ 
with real constants $u_1$, $u_2$, $v_1$, $v_2$. Therefore we have
$$f(z)=u(x,y)+iv(x,y)=(u_1+iv_1){z+\bar z\over 2}+(u_2+iv_2){z-\bar z\over 2i}+o\bigl(|z|\bigr)\qquad (z\to0)$$
or
$$f(z)=az+b\bar z+o\bigl(|z|\bigr)\qquad (z\to0)$$
with certain coefficients $a$, $b\in{\mathbb C}$. 
If $a=b=0$ then $f$ is complex differentiable at $0$ trivially. Otherwise  assume $a\ne0$ and  write
$$f(z)=a\bigl(z+c\bar z\bigr)+o\bigl(|z|\bigr)\qquad (z\to0)\tag{1}$$
with $c:={b\over a}$.
Let $z=re^{i\phi}$ with $r>0$. Then
$${f(z)\over z}=a\bigl(1+ce^{-2i\phi}\bigr)$$
and therefore
$$\arg{f(z)\over z}=\arg(a)+\arg(1+ce^{-2i\phi})\qquad(r>0)\ .$$
It follows that the existence of $\lim_{z\to0}\arg{f(z)\over z}$ necessitates $c=0$. From $(1)$ we then conclude that
$$f(z)=a z+o\bigl(|z|\bigr)\qquad (z\to0)\ ,$$
which means that $f$ is complex differentiable at $0$. The case $a=0$, $b\ne0$ is dealt with similarly; it is incompatible with the existence of $\lim_{z\to0}\arg{f(z)\over z}$.
