Showing the following function is entire.

Question

Say i have the following

$$\overline{f(\overline{x+iy})} = u(x,-y) - i v(x,-y)$$

How would i show using the Cauchy Riemann equations that the above is entire, would i use the chain rule to differentiate $u$ and $v$?.

Answer 1

Lets call the function $g(z):=\overline{f(\overline z)}$. There is another approach: If $f$ is holomorphic $\iff$ for every $z,z_0 \in \mathbb {C}$ there exists a continuous(!) function $\Delta(z)$ such that $$ f(z)=f(z_0)+\Delta(z)(z-z_0) $$ Complex conjugation is continuous and that makes the whole calculation easier. Overall we get, since f is entire, that for every $z,z_0 \in \mathbb {C}$ : $$ f(\overline{z})=f(\overline{z_0})+\Delta(\overline z)(\overline{z}-\overline{z_0}) $$ Conjugating again we get that: $$ g(z)=\overline{f(\overline{z})}=\overline{f(\overline{z_0})}+\overline{\Delta(\overline z)}(\overline{\overline z- \overline {z_0}})=g(z_0)+\overline{\Delta(\overline z)}(z-z_0) $$ Now the function $\overline{\Delta(\overline z)}$ is continuous as a composition of continuous functions.
Using Cauchy-Riemann is an option. It is more or less applying the chainrule, Ill give you a quick example: $$ \partial_y u(x,-y)=u_y(x,-y)(-1)=-u_y(x-y) \\ \partial_x v(x,y)=v_x(x,-y) $$ From that we get that, since $f$ is holomorphic, that $$ -u_y(x-y)-v_x(x,-y)=-u_y(x-y)-(-u_y(x,-y))=0 $$ Remember to check the second equation and that $g$ is totally differentiable as a real function!
Another notable thing is: The domain is $\mathbb{C}$. So if our function is defined and holomorphic for a $z \in \mathbb{C}$ it is also defined and holomorphic for $\overline z \in \mathbb{C}$. This may seem trivial here, but for more general domains, it might be tricky!