Given the function $f(z)=\text{Re}(z)+i |\text{Im}(z)|$. I need to check the complex differentiability again. I did a few of these kind of checks allready and they were not hard, but I have problems with this one. I want to use the definition only, no Cauchy-Riemann. Let $z=x+iy$: $$ \lim_{z\rightarrow z_0} \frac{f(z)-f(z_0)}{z-z_0} = \lim_{z\rightarrow z_0} \frac{x+i|y|-x_0-i|y_0|}{(x+iy)-(x_0+iy_0)} = \lim_{z\rightarrow z_0} \frac{(x-x_0)+i(|y|-|y_0|)}{(x-x_0)+i(y-y_0)} $$
Splitting this
i) $z_1= x_0+iy$ with $y\rightarrow y_0$ (vertical)
ii) $z_2=x+iy_0$ with $x\rightarrow x_0$ (horizontal)
We get:
i)$$ \lim_{z\rightarrow z_0} \frac{f(z_1)-f(z_0)}{z-z_0} = \lim_{y\rightarrow y_0} \frac{i(|y|-|y_0|)}{i(y-y_0)}=\lim_{y\rightarrow y_0} \frac{|y|-|y_0|}{y-y_0}$$
ii) The same way we get $$ \lim_{z\rightarrow z_0} \frac{f(z_2)-f(z_0)}{z-z_0} = \lim_{x\rightarrow x_0} \frac{x-x_0}{x-x_0}=1$$
Now it is only differentiable if $$ 1=\lim_{y\rightarrow y_0} \frac{|y|-|y_0|}{y-y_0} \Leftrightarrow |y|-|y_0| = y-y_0 $$
Where do I go from here? I guess I need to go trough the cases of $y_0>0,y_0=0,y_0<0$?
I guess you probably know that $z\mapsto\bar{z}$ is not differentiable. Now, you can see that your function is actually the same as:
$$f(z)=\begin{cases}z&\text{Im}(z)\ge 0\\\bar{z}&\text{Im}(z)\le 0\end{cases}$$
so in the half-plane $\text{Im}(z)>0$ it is differentiable and in the half-plane $\text{Im}(z)<0$ it is not.
As for the line $\text{Im}(z)=0$ (the real axis), you may just want to take the limit $\frac{f(z+i\Delta z)-f(z)}{i\Delta z}=\frac{|\Delta z|}{\Delta z}=\begin{cases}1&\Delta z>0\\-1&\Delta z<0\end{cases}$ along the imaginary axis ($z$-real, $i\Delta z$ pure imaginary) and directly prove that it doesn't exist, so the function is not differentiable there either.
Result: the function is differentiable in the half-plane $\text{Im}(z)>0$.