# Prove that $z|z|$ is not holomorphic at any point

I'm trying to prove that the complex-valued function $f(z)=z|z|$ is not holomorphic at any point. I know $f_2(z)=|z|$ fails to be holomorphic at any point of $\mathbb C$. Then, the basic fact about product of holomorphic functions is not a good avenue. I mean, since $f_1(z)=z$ is holomorphic at $\mathbb C$, if $f_2$ was not holomorphic at only one point $a$, then this point could give us the answer, by a direct test. But it is not the case. So, I guess I have to find a point where $f$ is not holomorphic. But I could not get it. For example, using Cauchy-Riemann equations, I found out that $f$ is differentiable at $z=(0;0)$. I wonder if I'm ignoring some good result for it... I'm sorry if it is a trivial question, but I'd be glad if someone could give me a hint, please. Thanks in advance!

Consider the holomorphicity of $\frac{f}{f_1}$. That solves your problem everywhere except $z = 0$.
Another way is to check complex differentiability : $f$ is $\mathbb C$-differentiable at $0$, because $$\frac{h|h|-0}{h}=|h| \xrightarrow[h\to0]{}0$$ but if $z\ne0$, $$\frac{f(z+h)-f(z)}{h} = |z+h|+z\frac{|z+h|-|z|}{h}$$ First term has limit $|z|$, OK. For second term, it can be shown that it doesn't have limit. For example, if $h=\alpha z$, with $\alpha\in\mathbb R$ sufficiently small, $$\frac{|z+h|-|z|}{h} = \frac{|z|}{z}$$ but for $h=z(e^{i\theta}-1)$, with $\theta\in\mathbb R$ tending to $0$, $$\frac{|z+h|-|z|}{h} = 0$$ which gives us two very different limits.
Use the Cauchy-Riemann criteria. Let $f(z)=f(x+iy)$. Now let $u(x,y)=Re(f(x+iy))$ and $v(x,y)=Im(f(x+iy))$. Now check if the partial derivatives satisfy $u_x=v_y$ and $u_y=-v_x$. If this holds, and the partials are continuous, then the original function $f(z)$ is holomorphic everywhere except where the Cauchy-Riemann criteria is not satisfied. Note the condition holds nowhere so the function is not holomorphic at any point.