Is it possible for a function to be differentiable in the complex plane but not in the real plane? I asked the question in the title.  

Is it possible for a function to be differentiable in the complex plane but not in the real plane? 

Could you help me find some examples or explain how it is possible? 
 A: Assume that 
$$f:\quad z=x+iy\quad \mapsto\quad w=u+iv$$is complex differentiable at the point $z_0=x_0+iy_0$, and let
$${\bf f}:\quad(x,y)\mapsto(u,v)$$
be the corresponding ${\mathbb R}^2$-map. I claim that ${\bf f}$ is ${\mathbb R}$-differentiable at ${\bf z}_0=(x_0,y_0)$.
Proof. Let $f'(z_0)=:c=a+ib\in{\mathbb C}$. Then
$$f(z_0+Z)-f(z_0)=c\>Z+o\bigl(|Z|\bigr)\qquad(Z\to0\in{\mathbb C})\ .$$
Separating real and imaginary parts here gives
$$\eqalign{
u(x_0+X,v_0+Y)-u(x_0,y_0)&=aX-bY+o\bigl(\sqrt{U^2+V^2}\bigr)\cr
v(x_0+X,v_0+Y)-v(x_0,y_0)&=bX+aY+o\bigl(\sqrt{U^2+V^2}\bigr)\cr}\qquad\bigl((U,V)\to(0,0)\in{\mathbb R}^2\bigr)\ .$$
This can be written as
$${\bf f}({\bf z}_0+{\bf Z})-{\bf f}({\bf z}_0)=\left[\matrix{a&-b\cr b&a\cr}\right]\left[\matrix{X\cr Y\cr}\right]+o\bigl(|{\bf Z}|\bigr)\qquad({\bf Z}\to{\bf 0})$$
and proves
$$\bigl[d{\bf f}({\bf z}_0)\bigr]=\left[\matrix{a&-b\cr b&a\cr}\right]\ .\qquad\square$$
A: Is possible that the function is not differentiable across the real line, but is differentiable across any other direction in the complex plane.
There is no "real plane", so I take it as "real line".
Take the function $|x|$. It is not differentiable at $x=0$, but if you generate a surface by translating the function horizontally across the complex plane, it is constant across that direction, so it is differentiable in that direction, even on the tip of the function.

You can smooth it fast enough on any direction (except the real line), to make it differentiable on any direction, except on the real line.
For example, the triangle function can be approximated by a fourier series. The constant line would be a single term in the fourier series, and the real line would have all the terms in the series. If you generate the surface by increasing the number of terms with the angle, from 0 terms at 90° to $\infty$ terms at 0°, the resulting surface would be differentiable on any direction, except the real line.
