Use the Cauchy-Riemann equations to show that the function $$g(z) = \sin (\bar z)$$
is not analytic at any point of $\mathbb{C}$.
Here's as far as I got –
$$\sin \left(\frac{\bar z}{1}.\frac{z}{z}\right)
=\sin \left(\frac{|z|^2}{z}\right)
=\sin \left(\frac{x^2 + y^2}{x+iy}\right)$$
I can't see how to separate the real and imaginary parts so that I can apply the Cauchy-Riemann equations.
Best Answer
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.