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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?

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    $\begingroup$ 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. $\endgroup$
    – Martin R
    Feb 18, 2022 at 12:31

3 Answers 3

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No, you did not assume the correct hypothesis to prove the wanted result by contradiction.

To see the error, consider the following modification:

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.

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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 $z_0$.

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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 $0$.

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