Showing $\sin(\bar{z})$ is not analytic at any point of $\mathbb{C}$

Writing $f(z) = \sin\bar{z} = \sin(x - iy) = \sin x \cosh y - i \cos x \sinh y$, we have

$f(z) = u(x,y) + i v(x,y)$

where

$u(x,y) = \sin x \cosh y $ and $v(x,y) = - \cos x \sinh y $

If the Cauchy - Riemann equations $u_x = v_y$ , $u_y = -v_x$ are to hold , it is easy to see that it will hold only at the points $z = \frac{\pi}{2} + n\pi$, $n \in\mathbb{Z}$.

Clearly, then, there is no neighborhood of any point throughout which $f$ is analytic. Also note that these are the isolated points. A function is analytic when Cauchy-Riemann equations hold in an open set. So, we may conclude that $\sin\bar{z}$ is nowhere analytic. Similarly we can show that $\cos\bar{z}$ is nowhere analytic.

Note that $\sin(\overline z)=\overline{\sin(z)}$. Since $\sin$ is analytic, if $g$ is analytic on some open set, then so is the real-valued function $\sin+g$. But a nonconstant real-valued function cannot be analytic (e.g., use the CR-equations or the open mapping theorem).

More generally, if $f$ is a nonconstant analytic function on a connected open set, then $g(z)=f(\overline z)$ is not analytic on any open set. It is true that $\overline g$ is analytic, so if $g$ were analytic then the real-valued functions $\mathrm{Re}(g)=\frac{1}{2}(g+\overline g)$ and $\mathrm{Im}(g)=\frac{1}{2i}(g-\overline g)$ would be analytic, hence constant, which means that $g$ (and hence $f$) would be constant.

Furthermore, the only points at which the Cauchy-Riemann equations can be satisfied for $g$, being the complex conjugate of the analytic function $\overline g$, are at the zeros of the derivative of $\overline g$, which are isolated.