Show that h(z) = z bar (line over the z) is not analytic on any domain. Check the Cauchy Riemann equations to do this.
Similiar question on an upcoming exam and I'm not sure how to do it. Can someone please show the steps?
Thanks a lot!
$\displaystyle z=x+iy\mapsto \bar{z}=x-iy\equiv u(x,y)+iv(x,y)$. Then $\displaystyle u_x = 1$, $\displaystyle u_y=0$, $\displaystyle v_x = 0$, $\displaystyle v_y = -1$. The Cauchy Riemann equations are $\displaystyle u_x=v_y$ and $\displaystyle u_y=-v_x$. The former is never satisfied, so the function is not holomorphic on any domain.
I'll add a little known (to some, maybe) but useful theorem:
When a function is expressed in the form $\displaystyle f(z, \bar{z})$, the Cauchy-Riemann relation is $\displaystyle \frac{\partial f}{\partial \bar{z}} = 0$.
Posting of proof available upon request.
So it's trivial to see that $\displaystyle f(z) = \bar{z}$ is not analytic .....