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? 

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.
