Proving that a function is nowhere differentiable I'm just going over a bit of revision for an upcoming exam, and I just wanted to verify whether my working/argument was sound. 
I've been asked to show that $f(z) = \frac{1}{\overline{z}}$ is nowhere differentiable on $\mathbb{C} \setminus \{0\}$
I started by letting $z = re^{i\theta}$, with $\overline{z} = re^{-i\theta}$. Then, I have; $$f(z) = \frac{1}{re^{-i \theta}} = \frac{1}{r} \cdot e^{i\theta}$$
Then, I set; $$u = \frac{\cos(\theta)}{r}, v = \frac{\sin(\theta)}{r}$$
And used the Cauchy-Riemann equations for polar co-ordinates to show the following (I skipped out a lot of working, but I'm quite confident that I'm correct with it);
$$\frac{\partial u}{\partial r} = \frac{1}{r} \cdot \frac{\partial v}{\partial \theta} \Rightarrow \frac{-\cos(\theta)}{r^2} = \frac{\cos(\theta)}{r^2} \Rightarrow 0 = 2 \cdot\cos(\theta) \Rightarrow \theta = \frac{\pi}{2} + n\pi$$
$$\frac{\partial v}{\partial r} = \frac{-1}{r} \cdot \frac{\partial u}{\partial \theta} \Rightarrow \frac{-\sin(\theta)}{r^2} = \frac{\sin(\theta)}{r^2} \Rightarrow 0 = 2 \cdot\sin(\theta) \Rightarrow \theta = n\pi$$
(In both cases, $n \in \mathbb{Z}$)
Hence, for these equations to be satisfied, we must have $n\pi = \frac{\pi}{2} + n\pi$, which has no solutions, so $f(z)$ is nowhere differentiable.
Is that sound enough?? Or should I be using a limit approach??

If $f(z)$ is differentiable at $z_0$ and $f(z_0)\ne0$, then $1/f(z)$ is differentiable at $z_0$. In your case, if $1/\bar z$ were differentiable at some point, then so would $1/f=\bar z$. But $\bar z$ does not satisfy the Cauchy--Riemann equations. Contradiction.
