Finding analyticity

Question

Given that f is analytic, under what conditions is $g(z)=\overline{f(z)}$ analytic?

Does this explanation make sense? : $g'(z)=lim_{h\rightarrow 0} \dfrac{g(z+h)-g(z)}{h}=lim_{h\rightarrow 0}\dfrac{\overline{f(z+h)}-\overline{f(z)}}{h}=\overline{f'(z)}$

Therefore, g(z) is analytic when $g'(z)=\overline{f'(z)}$

Answer 1

Let $f(z) = u(x,y) + iv(x,y)$ analytic on a domain $D \subset \mathbb{C}$. Then $g(z) = \overline{f(z)} = u(x,y) - iv(x,y)$ is analytic in $D$ if and only if $u(x,y)$ and $-v(x,y)$ have continuous first partial derivatives on $D$ that satisfy the Cauchy-Riemann equations. Thus, $g(z) = \overline{f(z)}$ must satisfy

$$\frac{\partial{u}}{\partial{x}} = -\frac{\partial{v}}{\partial{y}} \;\; \text{and} \;\; \frac{\partial{u}}{\partial{y}} = \frac{\partial{v}}{\partial{x}}.$$

Answer 2

Consider the Cauchy-Riemann equations, Rudin Real and Complex Analysis 11.2. If $f=u+iv$ with $u$, $v$ real, they require for analyticity that $u_x=v_y$ and $u_y=-v_x$ where the subscripts denote differentiation. For $g=u-iv$, the Cauchy-Riemann equations required for $g$ to be analytic are $u_x=-v_y$ and $u_y=v_x$. It would seem that all four of these equations can be met only if $v_x=0$ and $v_y=0$, which in turn implies $u_x=0$ and $u_y=0$, i.e., $f$ constant.

In your solution, $$\lim_{h\rightarrow 0}\dfrac{\overline{f(z+h)}-\overline{f(z)}}{h}=\overline{f'(z)}$$ is not correct, because the limit is $d\overline{f(z)}/dz$, and that is not shown to equal the complex conjugate of $f'(z)$.