Prove that $f(z) = |z|$ is not analytic.

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.