Analytic complex function 
Show that $f(z) = \bar{z}$ is nowhere differentiable.

I must use the definition in order to do this, the definition is:
$$\frac{df}{dz}(z_0) = f'(z_0) = \lim_{\triangle z \rightarrow 0}\frac{f(z_0+\triangle z)-f(z_0)}{\triangle z}$$ thus I will have 
$$\frac{\bar{(z_0} + {\bar\triangle z)}-{\bar{z_0}}}{\bar\triangle z} = \frac{\bar{\triangle z}}{\bar\triangle z}$$
But then what further process do I have to do in order to show it is not differentiable? 
 A: Write $\Delta z = re^{i\theta}$. Then the derivative is
$$\lim\limits_{r\to 0}\frac{\overline{z+re^{i\theta}}-\bar z}{re^{i\theta}}=\lim_{r\to 0}\frac{\overline{re^{i\theta}}}{re^{i\theta}}=\lim_{r\to 0}\frac{re^{-i\theta}}{re^{i\theta}}=\lim_{r\to 0}e^{-2i\theta}=e^{-2i\theta}$$
which is not constant in $\theta$, since (for example) $e^{-2i0}=1$ and $e^{-2i(\pi/2)}=-1$.
A: Pick any $z = x + i y$. And consider two sequences that both converge to $z$ as $n \to \infty$:


*

*$z_n = (1/n + x) + i y$


$$
\lim_{n \to \infty} \dfrac{\overline{z_n} - \overline{z}}{z_n - z} = \lim_{n \to \infty} \dfrac{1/n}{1/n} = 1
$$


*

*$z_n = x + i(1/n + y)$


$$
\lim_{n \to \infty} \dfrac{\overline{z_n} - \overline{z}}{z_n - z} = \lim_{n \to \infty} \dfrac{-1/n}{1/n} = -1
$$
Since there are two sequences that make the quotient converge to different limits, the limit $\lim_{w \to z}\dfrac{\overline{w} - \overline{z}}{w - z}$ doesn't exist. Hence, the function isn't differentiable.
A: By the definition,
$$
\begin{align}
\frac{\mathrm{d}\overline{z}}{\mathrm{d}z}
&=\lim_{h\to0}\frac{\overline{z+h}-\overline{z}}{h}\\
&=\lim_{h\to0}\frac{\overline{h}}{h}\\
&=\lim_{h\to0}\frac{r_he^{-i\theta_h}}{r_he^{i\theta_h}}\\
&=\color{#C00000}{\lim_{h\to0}e^{-i2\theta_h}}
\end{align}
$$
For the limit to exist, the limit in red must exist. However, $\theta_h$ can be anything while $h\to0$.
A: You conjugated the denominator when you didn't need to.but other than that, you should see what happens in the limit if $\Delta z$ is entirely imaginary, or entirely real.
