Prove that $f(z) = |z|$ is not analytic. I do not fully grasp what it means to be analytic and therefore do not know the conditions on which I have to show in order to complete the proof.
 A: If a function is analytic in a neighborhood of a point, then it satisfies the Cauchy-Riemann equations.
But we have 
$$|z|=\sqrt{x^2+y^2}$$
Clearly, with $u=\sqrt{x^2+y^2}$ and $v=0$ we see that for $(x,y)\ne (0,0)$
$$\frac{\partial u}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}\ne 0 =\frac{\partial v}{\partial y}$$
And for $z=0$, we see that $\lim_{\Delta \to 0}\frac{|0+\Delta z|-|0|}{\Delta z}$ fails to exist.
Therefore, $|z|$ is nowhere analytic.
A: Every analytic function is differentiable. But $f$ isn't, that is, the limit$$\lim_{z\to0}\frac{|z|}z$$does not exist (as in the reals). So, $f$ is not analytic.
A: Given $z, a, t\ne0$, with $t>0$ real, we have
\begin{eqnarray}
\lim_{t\to0}\frac{|z+at|-|z|}{at}&=&\lim_{t\to0}\frac{|z+at|^2-|z|^2}{at(|z+at|+|z|)}=\lim_{t\to0}\frac{t(\bar{a}z+a\bar{z})+a^2t^2}{t(|z+at|+|z|)}\\&=&
\lim_{t\to0}\frac{\bar{a}z+a\bar{z}+at}{|z+at|+|z|}=\frac{\bar{a}z+a\bar{z}}{2|z|}
\end{eqnarray}
this shows two things: first, $f$ is not holomorphic for all $z\ne0$, and second if it were holomorphic it would have been at $z=0$. Since we also have
$$
\lim_{t\to0}\frac{|0+at|-|0|}{at}=\lim_{t\to0}\frac{|a|t
}{at}=\frac{|a|}{a}
$$
Hence $f$ is not holomorphic.
