Is $sin(frac{1}{|z|})$ holomorphic on $Bbb C-{0}$?

Question

$f(z)=sin(frac{1}{|z|})$, $zin Bbb C-{0}$.
$$frac{partial f}{partialbar{z}}=frac{partial sin(frac{1}{|z|})}{partialbar{z}}=frac{partial}{partial bar{z}}sin(frac{1}{(zbar z)^{1/2}})=cos(frac{1}{|z|})(2^{-1}z)(zbar{z})^{frac{-3}{2}} neq 0$$
So, $f(z)$ is not holomorphic on $Bbb C-{0}$.
Am I correct?

Fred · Answer

More general: if $D$ is a domain in $ mathbb C$ and if $f:D to mathbb R$ is real valued, then $f$ is holomorphic on $D iff$ $f$ is constant.
Cauchy-Riemann !

José Carlos Santos · Answer

Yes, that is correct. Another way of proving it is this: if $z=x+yi$, with $x,yinBbb R$; then write $f(x+yi)$ as $u(x,y)+v(x,y)i$. Then $v$ is the null function. So, if $f$ was holomorphic, you would have (by the Cauchy-Riemann equations) $u_x=u_y=0$ and s $u$ would be constant. But then $f$ itself would be constant. But it is not.

Answered by José Carlos Santos on December 8, 2021

Is $sin(frac{1}{|z|})$ holomorphic on $Bbb C-{0}$?

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