Which points $\frac{1}{z \bar{z}}$ is it holomorphic? I am currently looking for each point $\frac{1}{z \bar{z}}$ is holomorphic. 
Could I use the fact that $\frac{1}{z}$ is holomorphic except for the point $z=0$. Do I have to use Cauchy-Riemann theorem to solve it?
 A: We can show that $f(z)=\frac1{z\bar z}$ is nowhere analytic without appealing to the Cauchy-Riemann Equations.  To proceed, we find that the difference quotient $\frac{\Delta f(z)}{\Delta z}$ can be written as
$$\begin{align}
\frac{f(z+\Delta z)-f(z)}{\Delta z}&=\frac{\frac{1}{|z+\Delta z|^2}-\frac1{|z|^2}}{\Delta z}\\\\
&=\left(\frac{|z|^2-|z+\Delta z|^2}{\Delta z}\right)\left(\frac{1}{|z|^2\,|z+\Delta z|^2}\right)\\\\
&=\left(\frac{-|\Delta z|^2-\bar z\Delta z-z\overline{\Delta z}}{\Delta z}\right)\left(\frac{1}{|z|^2\,|z+\Delta z|^2}\right)\\\\
\end{align}$$
It is easy to see that $\lim_{\Delta z\to 0}\frac{|\Delta z|^2}{\Delta z}=0$, and $\lim_{\Delta z\to 0}\frac{\bar z\Delta z}{\Delta z}=\bar z$.  
However, $\lim_{\Delta z\to 0}\frac{\overline {\Delta z}}{\Delta z}$ does not exist since if the limit is approached along $\Delta z=\Delta x\to 0$, then 
$$\lim_{\Delta z\to 0}\frac{\overline {\Delta z}}{\Delta z}=\lim_{\Delta x\to 0}\frac{\Delta x}{\Delta x}=1 \tag 1$$
whereas if the limit is approached along $\Delta z=i\Delta y \to 0$, then 
$$\lim_{\Delta z\to 0}\frac{\overline {\Delta z}}{\Delta z}=\lim_{\Delta y\to 0}\frac{-i\Delta y}{i\Delta y}=-1 \tag 2$$
Since the limits in $(1)$ and $(2)$ are not equal, then the limit $\lim_{\Delta z\to 0}\frac{\overline {\Delta z}}{\Delta z}$ does not exist.
Hence, we conclude $f$ is nowhere differentiable and thus nowhere analytic.
