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? 

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 $e^{-2i(\pi/2)}=-1$.
