Show that the function $f(z)=\bar{z}, z \in \mathbb{C},$ is not complex differentiable at $0$, but is differentiable in the real sense at all points in $\mathbb{C}$.
I am trying to brush up on some old complex analysis notes and I can't seem to figure this proof out. I am guessing that "in the real sense" is referring to real differential numbers. From here on I am stumped, so any guidance would be appreciated.
On
It is not complex differentiable, or holomorphic because the limit
$$\lim\limits_{z \to z_0} \frac{\bar{z} - \bar{z_0}}{z-z_0}$$
Does not exist for any $z_0 \in \mathbb{C}$. However we can interpret it as the map $\mathbb{R}^2 \to \mathbb{R}^2$ sending $(x,y)$ to $(x,-y)$ which certainly has a total derivative in the real multivariable sense. In particular, since this map is linear it is it's own total derivative at any point.
On
In the real sence this will be: $$\lim_{(x,y)\to(x_0,y_0)}\frac{(x,-y)-(x_0,-y_0)}{\Vert (x-x_0,y-y_0)\Vert}$$ $$=\lim_{(x,y)\to(x_0,y_0)}\frac{(x-x_0,-y+y_0)}{\Vert (x-x_0,y-y_0)\Vert}$$ $$=\lim_{(x,y)\to(x_0,y_0)}\frac{(x-x_0,0)}{\Vert (x-x_0,y-y_0)\Vert}-\lim_{(x,y)\to(x_0,y_0)}\frac{(0,y-y_0)}{\Vert (x-x_0,y-y_0)\Vert}=0$$
On
If it were complex differentiable, then it would satisfy the Cauchy-Riemann equations. So if we put $z=x+iy$, then
$$f(z)=x-iy.$$
Where $u(x,y)=x,v(x,y)=-y.$ By C-R equations, we have
$$u_x=v_y, u_y=-v_x.$$
So $u_x=1, v_y=-1$ and by C-R we'd arrive at $$-1=1$$
which is absurd!! Hence, $f(z)=\bar{z}$ is not complex differentiable.
I might be wrong, but I don't think that's differentiable anywhere on the complex plane.
The equation doesn't satisfy the CR equations.