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

Question

$f(z)=\sin(\frac{1}{|z|})$, $z\in \Bbb C-\{0\}$.

$$\frac{\partial f}{\partial\bar{z}}=\frac{\partial \sin(\frac{1}{|z|})}{\partial\bar{z}}=\frac{\partial}{\partial \bar{z}}\sin(\frac{1}{(z\bar z)^{1/2}})=\cos(\frac{1}{|z|})(2^{-1}z)(z\bar{z})^{\frac{-3}{2}} \neq 0$$

So, $f(z)$ is not holomorphic on $\Bbb C-\{0\}$.

Am I correct?

Answer 1

Yes, that is correct. Another way of proving it is this: if $z=x+yi$, with $x,y\in\Bbb R$; 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.

Answer 2

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 !