# Can the product rule be used to prove that $z\vert{z}\vert$ is not holomorphic at any point?

Proposition 5.8.(1):

Suppose that $$f,g:U\longrightarrow\mathbb{C}$$ are holomorphic. Then $$f\cdot g:U\longrightarrow\mathbb{C}$$ are holomorphic and $$(f\cdot g)'=f'\cdot g+f\cdot g'$$.

In order to show that $$f(z)=z\vert{z}\vert$$ is not holomorphic anywhere, assume for contradiction that $$a(z)=z$$ and $$b(z)=\vert{z}\vert$$ both are holomorphic functions. This means that $$f=a\cdot b$$ is holomorphic and $$(a\cdot{b})'=a'\cdot b+a\cdot b'$$, but $$b(z)$$ is not holomorphic in any point and $$b'(z)$$ does not exist. Therefore $$f(z)$$ cannot be holomorphic at any point.

Could this pass for a proof or do I need to pull out the Cauchy-Riemann equations?

• In order to show that $f(z)=z\vert{z}\vert$ is not holomorphic you must assume that f is holomorphic and derive a contradiction. Feb 18, 2022 at 12:31

In order to show that $$f(z)=z^2$$ is not holomorphic anywhere, assume for contradiction that $$a(z)=z(1+|z|)$$ and $$b(z)=z/(1+|z|)$$ both are holomorphic functions. This means that $$f=a\cdot b$$ is holomorphic and $$(a\cdot{b})'=a'\cdot b+a\cdot b'$$, but $$b(z)$$ is not holomorphic in any point and $$b'(z)$$ does not exist. Therefore $$f(z)$$ cannot be holomorphic at any point.
No, that does not work. The theorem that you mentioned says that if $$a$$ and $$b$$ are differentiable at $$z_0$$, then $$f$$ is also differentiable at $$z_0$$. It does not tell you what happens if $$a$$ or $$b$$ is not differentiable at $$z_0$$.
The reasoning you wrote does not work. In order to apply the formula $$(a.b)'=a'b+ab'$$, you have to have that both f and g are holomorphic, which is what you are trying to prove to get your contradiction. Here is a simple way you could prove it without using Cauchy-Rieman equations:
Assume that $$f(z)=z|z|$$ is a holomorphic function. Then it has a Taylor expansion at $$0$$ and can be written locally as $$f(z)=\sum \limits_{n \in \mathbb{N}} a_n z^n$$. Since $$f(0)=0$$, then $$a_0=0$$ and so $$\frac{f(z)}{z}=\sum \limits_{n \in \mathbb{N}}a_{n+1}z^n$$ and since $$\frac{f(z)}{z}=|z|$$, that would prove that $$|z|$$ a holomorphic function, which is not the case since as a real valued function, it is not differentiable at $$0$$.