Ytukyg

1. By definition the derivative of $overline{z}$ is:

begin{align}
lim_{Delta hto 0}frac{overline{z+Delta h}-overline{z}}{Delta h}&=lim_{Delta hto 0}frac{overline{z}+overline{Delta h}-overline{z}}{Delta h}
\&=lim_{Delta hto 0}frac{overline{Delta h}}{Delta h}
\&=lim_{Delta hto 0}frac{{Delta x-iDelta y}}{Delta x+iDelta y}
\&=lim_{Delta hto 0}frac{r(cosalpha-isinalpha)}{r(cosalpha+isinalpha)}
\&=lim_{Delta hto 0}frac{cosalpha-isinalpha}{cosalpha+isinalpha}
end{align}

How should it be continue with turning over to $frac{e^{-ialpha}}{e^{ialpha}}$?

2. Is there an intuitive explanation why the conjugate function is not differentiable? As it is just "reflection" on $x$- axis

Intuitively, a function $f$ of one variable is differentiable at a point $p$ if for some tiny region $U$ around $p$, applying $f$ looks like multiplying by some number / scalar (in other words, scaling and rotating), relative to $p$ and $f(p)$. No matter how you scale and rotate, you can never flip. Thus complex conjugation doesn't look like scalar multiplication locally, and therefore is not differentiable.

This will not expand on your limit proof in your question. But there is this nice theorem in complex analysis, called the identity theorem:

Theorem. Given two complex differentiable functions $f$ and $g$ and a set $D$ with some limit point (e.g. an open set, a line, ...). If $f(z)=g(z)$ for all $zin D$ then we already have $f=g$.

Your function $f(z)=z$ and its conjugate $bar f$ agree on the real line, i.e. $Bbb RsubsetBbb C$. The real line is a set with limit point, so when both functions are complex differentiable, then $z=bar z$ for all $zinBbb C$. Obviously wrong. But because we know that $z$ is complex differentiable, $bar z$ can not.

This works for all non-constant functions $f$ for which the restriction to $Bbb R $ is real, i.e. $f|_Bbb R:Bbb RtoBbb R$. For example, this shows that $overlinesin (z)$ is not complex differentiable.

You can see that by approaching to the origin first on the real axis and second on the complex axis, that means: Consider for the first limit $Delta x=0$ and for the second limit consider $Delta y=0$. You get two different limits, and therefore the limit does not exist. Hence $f(z)=overline{z}$ is not differentiable.

Conjugation is real-differentiable, in the sense that all partial derivatives of it exist everywhere. However, this does not mean it's complex differentiable. A real differentiable function $x+yimapsto u(x,y)+iv(x,y)$ is complex differentiable provided that
$$begin{align}frac{partial u}{partial x} &= frac{partial v}{partial y}\
frac{partial u}{partial y} &= -frac{partial v}{partial x}end{align}$$
These are the Cauhcy-Riemann eComplex conjugation does not satisfy this, since $u(x,y)=x, v(x,y)=-y$.

Intuitively, $zmapsto overline z$ is not differentiable since its directional derivatives are different depending on their direction, as opposed to a function like $z^2$.