If $f(z),\overline {f(z)}$ are both analytic on a disc $D$ then what can be said about $f$ on $D$ is constant? Thank you.
What I'm thinking is:
$$f(z)=u(x,y)+iv(x,y), \quad \bar{f}(z)=u(x,y)-iv(x,y)$$
Let $f(z) = u(x,y) + i v(x,y)$, then $\bar f(z) = u(x,y) - iv(x,y)$.
- Cauchy-Riemann equations for $f$: $(1) \quad \partial_xu = \partial_yv$, $(2) \quad \partial_yu = -\partial_xv$
- Cauchy-Riemann equations for $\bar f$: $(3) \quad \partial_xu = -\partial_yv$, $(4) \quad \partial_yu = \partial_xv$
Combine $(1)+(3)$ to get $\partial_xu = \partial_yv = - \partial_xu$, therefore $\partial_xu = \partial_yv = 0$. Similarly, $(2)+(4)$ imply that $\partial_yu = \partial_xv = 0$. $u$ and $v$ have all their partial derivatives vanishing, D is connected (being a domain), therefore $u$ and $v$ are constant, and $f$ is a constant on D? Would I be right with this?