$f(|z|)$ is not an analytic function

Question

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.

Answer 1

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.

Answer 2

You are very close to the solution. Choose $z=x+iy$ on a circle with radius $r$: $|z|=r$. Then you get $$ x\frac{\partial u}{\partial r}(r) = y\frac{\partial v}{\partial r}(r) $$ If $z$ runs over the circle the derivatives in the equation above are always the same, but $x$ and $y$ vary. If you choose $x=r,y=0$ and $x=0,y=r$ you get $$ \frac{\partial u}{\partial r}(r) = \frac{\partial v}{\partial r}(r) = 0 $$ Hence $u=const.$ and $v=const.$ and $f$ must be constant.

Answer 3

Assume that $g$ is holomorphic. Fix a $z\ne0$ and consider the auxiliary function $$\psi(t):=g\bigl(e^{it}z\bigr)\qquad(t\in{\mathbb R})\ ,$$ which is in fact constant. It follows that $$\psi'(t)=g'\bigl(e^{it}z\bigr)\>ize^{it}\equiv0\ ,$$ and putting $t=0$ here implies $g'(z)=0$. It would follow that $$f'(x)=g'(x)\equiv0\qquad(x>0)\ ,$$ contrary to assumption about $f$.