How to show that $f(z) = \bar{z}$ is not differantiable in the complex plane? I know that $\bar{z} = \frac{1}{z}=e^{-i\theta}=\cos \theta - i \sin \theta$
We also know that $z+dz = (x+\delta x) + i(y + \delta y)$, therefore $\overline{z+dz} = 1/ ((x+\delta x) + i(y + \delta y))$
and therefore all we have to do is to show that two paths lead to a different result.
Using the definition of the derivative we have that:
$$f'(z)=\lim_{\delta z \rightarrow 0}\frac{\overline{z+\delta z}-\bar{z}}{\delta x + i \delta y}$$
Am I on the right track? Because if all of the above is correct, then I am having trouble simplifying this.
 A: Not like that. Write $z=z_0+h$ with $h=re^{i\theta}$ and note that the limit
$$
\lim_{z\to z_0}\frac{\overline z-\overline z_0}{z-z_0}=\lim_{h\to0}\frac{\overline h}h=\lim_{h\to0}e^{-2i\theta}
$$
does not exist, since it would give different values for different values of $\theta$.
A: use the cauchy riemann equtaion in polar form $$\frac {\partial f}{\partial \bar z}=0$$ what you get is $$1=0$$ and that is never true or absurd hence the function is not differentiable in the entire complex plane
A: For my own reference:
$\overline{z+dz}$ is conjugate of $z+dz$, which is $(x+\delta x) + i(y+\delta y)$. Therefore,$\overline{z+dz}= (x+\delta x) - i(y+\delta y)$.
Hence: $$\overline{z+dz}-\bar{z} = (x+\delta x) - i(y+\delta y) - x - iy = \delta x - i\delta y$$
Thus we have
$$\lim_{\delta z \rightarrow 0} \frac{\delta x - i \delta y}{\delta x + i \delta y}$$
When we make $\delta x \rightarrow 0$ we get $-1$.\
When we make $\delta y \rightarrow 0$ we get $1$.
Thus, $\bar{z}$ is not differentiable everywhere.
