$f(|z|)$ is not an analytic function Let $f: [0,\infty)\rightarrow \mathbb{C}$ is a non constant function. Define $g:\mathbb{C}\rightarrow\mathbb{C}$ by $g(z)=f(|z|)$. Prove that $g(z)$ is not holomorphic.
So, I need to find a point $z_0$ where it is not analytic.  
Write $f(x)=u(x)+iv(x)$ then $f(|z|)=u(\sqrt{x^2+y^2})+iv(\sqrt{x^2+y^2})$. 
Set $p(x,y)=u(\sqrt{x^2+y^2})$ and $q(x,y)=v(\sqrt{x^2+y^2})$
Then, $$\partial p/\partial x=(\partial u/\partial x)(x/\sqrt{x^2+y^2}),\partial p/\partial y=(\partial u/\partial y)(y/\sqrt{x^2+y^2})$$
and  $$\partial q/\partial x=(\partial v/\partial x)(x/\sqrt{x^2+y^2}),\partial q/\partial y=(\partial v/\partial y)(y/\sqrt{x^2+y^2})$$
Suppose it is holomorphic then it has to satisfy cauchy riemann equations and so we should have
$$x(\partial u/\partial x)=y(\partial v/\partial y),y(\partial u/\partial y)=-x(\partial v/\partial x)$$
After writing this then i realized that $u,v$ are functions of single variable.. It does not make sense in trying to differentiate with two variables...
I am clueless at this stage.

Here's a different approach that uses a little bit of complex analysis. Suppose that $g$ is entire (analytic on $\mathbb{C}$). Then on the unit circle, $g$ is constantly equal to $f(1)$. As the unit circle has a limit point in $\mathbb{C}$, and since $\mathbb{C}$ is connected, $g(z)=f(1)$ for all $z \in \mathbb{C}$. This implies $f(t)=g(t)=f(1)$ for all $t \geq 0$, so $f$ is constant.
