Cauchy-Riemann equations for $f(z)=\overline z$ Show that the function $f(z) = \overline{z}$
is Not differentiable anywhere in the $z$-plane.
I am thinking about Cauchy-Riemann theorem $ f(z) = x-iy $
Here
$$
\begin{cases}
u(x,y) = x \\
v(x,y) = -y
\end{cases}
$$
and both of them are differentiable and 
\begin{align}
\frac{\partial u}{\partial x} = 1 \ne
 \frac{\partial v}{\partial y} \\
\frac{\partial u}{\partial y} = 0 = \frac{\partial v}{\partial x}
\end{align}
I really can't give an example to prove that the function isn't differentiable anywhere.
 A: In your example, $v(x,y) = -y$, not $y$.
A: The fact that the Cauchy Riemann equation isn't verified is quite enough.
For a direct proof,
$$f'(z):=\lim_{h\to0}\frac{\overline{z+h}-\overline z}h={}\lim_{\epsilon\to0}\frac{\overline{z+\epsilon e^{i\theta}}-\overline z}{\epsilon e^{i\theta}}=\frac{e^{-i\theta}}{e^{i\theta}}=e^{-i2\theta},$$ which depends on $\theta$.
A: You could also use the definition of a complex derivative and let $h$ go to $0$ on the real axis, and then on the imaginary axis:


*

*For $h  \space \epsilon \space \mathbb{R}$ is $\frac{f(z + h) - f(z)}{h}$ equal to $\frac{2h-h}{h} = 1$

*For $h \space \epsilon \space \mathbb{C}$  is $\frac{f(z + h) - f(z)}{h}$  equal to $\frac{- 2ih +ih}{ih} = -1$
Contradiction.
A: We have $f(z)=\bar z$
\begin{align}
f'(z)&=\lim_{{\Delta z}\to0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z}\\
&=\lim_{{\Delta z}\to0}\frac{\overline {z_0+\Delta z}-\overline z_0}{\Delta z}\\
&=\lim_{{\Delta z}\to0}\frac{\overline z_0+\overline\Delta z-\overline z_0}{\Delta z}\\
&=\lim_{{\Delta z}\to0}\frac{\overline\Delta z}{\Delta z}\\
&\\
&\text{If $\Delta z\to0$ through $(\Delta x,0)$,then}\\
&\\
f'(z)&=\lim_{(\Delta x,0)\to(0,0)}\frac{\Delta x}{\Delta x}=1\\
&\\
&\text{If $\Delta z\to0$ through $(0,\Delta y)$,then}\\
&\\
f'(z)&=\lim_{(0,\Delta y)\to(0,0)}\frac{-i\Delta y}{i\Delta y}=-1\\
\end{align}
i.e $f(z)=\bar z$ is nowhere differentiable in complex plane.
