If $Ref, Imf \in C^1$ at $z_0$, $\lim_{z\rightarrow z_0}Re\Big(\frac{f(z)-f(z_0)}{z-z_0}\Big)\in \mathbb{R}$, prove $f$ differentiable at $z_0$. 
If $f:A\rightarrow \mathbb{C}$ is such that $Ref, Imf$ are $C^1$ at $z_0\in A$ and the limit:
  $$\lim_{z\rightarrow z_0}Re\bigg(\frac{f(z)-f(z_0)}{z-z_0}\bigg)$$
  exists in $\mathbb{R}$, prove that $f$ is differentiable at $z_0.$

Attempt. All we need to prove is that 
$$\lim_{z\rightarrow z_0}Im\bigg(\frac{f(z)-f(z_0)}{z-z_0}\bigg)$$
also exists in $\mathbb{R}.$ Writing 
$$\frac{f(z)-f(z_0)}{z-z_0}=\frac{u(z)+iv(z)-u(z_0)-iv(z_0)}{x-x_0+i(y-y_0)}$$ didn't get me somewhere.
Thank you in advance.
 A: We may assume $z_0=f(z_0)=0$. By assumption
$$f(z)= u(x,y)+ i v(x,y)=(ax+by)+i(cx+dy)+o\bigl(|z|\bigr)\quad(z\to0)$$
with $a$, $b$, $c$, $d\in{\mathbb R}$. It follows that
$${f(z)\over z}={f(z)\>\bar z\over|z|^2}={(ax+by)(x-iy)+i(cx+dy)(x-iy)\over|z|^2}+o(1)\quad(z\to0)$$
and therefore
$${\rm Re}{f(z)\over z}={ax^2+dy^2 +(b+c)xy\over x^2+y^2}+o(1)\quad(z\to0)\ .$$
It has been shown many times here that the fraction on the RHS has a limit for $(x,y)\to(0,0)\>$ iff $\>a=d$ and $b+c=0$. But this means that $u$ and $v$ together satisfy the CR equations at $(0,0)$.
A: Suppose $\underset{z\rightarrow z_0}{lim}\frac{f(z)-f(z_0)}{z-z_0}=a+bi$
$\underset{z\rightarrow z_0}{lim}\frac{f(z)-f(z_0)}{z-z_0}=\underset{z\rightarrow z_0}{lim} \frac{u(z)+i v(z)-u(z_0)-i v(z_0) }{z-z_0}=
\underset{z\rightarrow z_0}{lim}\frac{u(z)-u(z_0)}{z-z_0}+ i(\underset{z\rightarrow z_0}{lim}\frac{ v(z)-(z_0)}{z-z_0})\Rightarrow $
$\underset{z\rightarrow z_0}{lim}\frac{u(z)-u(z_0)}{z-z_0}=a$ and $\underset{z\rightarrow z_0}{lim}\frac{ v(z)-(z_0)}{z-z_0}=b$
