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Show that $\frac{z}{1+|z|}$ is not holomorphic anywhere? How to show that without using C-R equations?

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If $f(z) =\frac{z}{1+|z|}$ is holomorphic on $U$, since $f(z) \ne 0$ on $\mathbb{C} \setminus \{0\}$ and $z$ is holomorphic, then $\left(\frac{z}{f(z)}-1\right)^2 = |z|^2$ is holomorphic on $U \setminus \{0\}$, an obvious contradiction since $|z|^2$ is complex-differentiable only at $z=0$

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    So you need to show that $g(z),h(z) \ne 0$ holomorphic on $U \implies g(z)+h(z), g(z)h(z), \frac{g(z)}{h(z)}$ are holomorphic on $U$ – 2017-02-05
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    We can show that $(\frac{z}{f(z)}-1)=|z|$ is holomorphic on $U\{0}$ which leads us to contradiction since $|z|$ is also differentiable only at z=0. Therefore, f(z) is not holomorphic on U. – 2017-02-06
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Note that $$\frac{z}{1+|z|}=\frac{z}{1+\sqrt{z\bar z}}$$

and $$\frac{\partial }{\partial \bar z}\left(\frac{z}{1+\sqrt{z\bar z}}\right)\ne 0, \,\,\text{for}\,\,z\ne 0$$

which implies that the Cauchy-Riemann equations are not satisfied (See this).

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    Why not, but how do you compute $\frac{\partial }{\partial \bar z}\left(\frac{z}{1+\sqrt{z\bar z}}\right)$ ? – 2017-02-05
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    @user1952009 [See this](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations#Independence_of_the_complex_conjugate). We view $z$ and $\bar z$ as independent. The CREs can be compactly written $$\frac{\partial f(z,\bar z)}{\partial \bar z}=0$$ – 2017-02-05
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    I mean it is not so obvious. So how do **you** compute it ? – 2017-02-05
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    @user1952009 $$\frac{\partial }{\partial \bar z}\left(\frac{z}{1+\sqrt{z\bar z}}\right)=-\frac{z^{3/2}}{2\sqrt{\bar z}(1+|z|)^2}$$ – 2017-02-05
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    So you are saying that we can replace syntaxically $z, \overline{z}$ by $x,y$ two real independent variables and differentiate with respect to $y$ ? In that case, it is a theorem, whose proof is not so obvious, at this level – 2017-02-05
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    @user1952009 Have you had a chance to look at the embedded link? – 2017-02-05
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    Yes, but the fact you didn't write a proof is the problem, revealing that the proof isn't so obvious. – 2017-02-05
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    @user1952009 Would I need to write a proof of the CREs? I have provided a reference for the well-known result. – 2017-02-05
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HINT try to calculate $\lim_{x->0+}\frac{f(ix)-f(o)}{ix}$ and $\lim_{x->0+}\frac{f(-x)-f(o)}{x}$

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    Thanks a lot. I wil try it – 2017-02-05