1. ## Approaching complex problem.....

Lets say we have a function: $f(z) = x+i|y|$.

We want to find all point that is it differentiable. Right of the bat we know that it is differentiable for $y >0$. For $f(z) = x+iy$ is a "direct" function of $x+iy$. E.g. we do not have to appeal to CR equations.

Also it is not differentiable for $y<0$ since $f(z) = x-iy$ is not a "direct" function of $x+iy$.

Now for $y = 0$ can we use the same argument to show that $f(z) = x+i|y|$ is not differentiable there?

Because then we have $f(z) = x$. And this is not a "direct" function of $x+iy$.

2. I mean analytic takes care of everything. Its stronger then differentiability. Thats what "function of z" is.

3. Originally Posted by heathrowjohnny
Lets say we have a function: $f(z) = x+i|y|$.

We want to find all point that is it differentiable. Right of the bat we know that it is differentiable for $y >0$. For $f(z) = x+iy$ is a "direct" function of $x+iy$. E.g. we do not have to appeal to CR equations.

Also it is not differentiable for $y<0$ since $f(z) = x-iy$ is not a "direct" function of $x+iy$.

Now for $y = 0$ can we use the same argument to show that $f(z) = x+i|y|$ is not differentiable there?

Because then we have $f(z) = x$. And this is not a "direct" function of $x+iy$.
For $y=0$ you can use the definition of differenciable, $\lim_{z\to z_0} \frac{f(z)-f(z_0)}{z-z_0}$. Now $z=x+iy$ and $z_0 = x$. Therefore, $\lim_{(x,y)\to (x_0,0)}\frac{x+i|y| - x}{iy} = \lim_{y\to 0}\frac{y}{|y|}$.
And this limit does not exist.

4. so you can't just say that $f(z) = x$ is not an "analytic function" of $x+iy$?