Proving that a Function is Analytic Given that it is Equal to the Complex Conjugate of an analytic Function

So I'm working a problem that states:

A function $f$ is analytic in an open set $U$. Define $g$ by $g(z)=\overline{f(\overline{z})}$ (just because the notation can be hard to read, this is the the complex conjugate of the function $f$ defined at the complex conjugate of $z$). Show that $g$ is analytic in the oopen set $U^{\star}=\{z:\overline{z}\in{U}\}$ and that $g^{\prime}(z)=\overline{f^{\prime}(\overline{z})}$ (this is the complex conjugate of the derivative of $f$ evaluated at the complex conjugate of $z$) for $z\in{U^{\star}}$

Now, my book gives the definition of an analytic function as one that is a function defined over an open set on which it is differentiable at every point of that set and that every derivative of that function satisfying the preceeding properties has derivatives that are analytic on that domain as well.

So, I know that being differentiable at a point means that a function is continuous at that point, so I see straight away that this means $\forall{\epsilon}>0$ there exists a $\delta>0$ such that if $|z-z_0|<\delta$ then $|f(z)-f(z_0)|<\epsilon$. Since $U$ is an open set, it follows that $U\cap{z_0}$ is non-empty. Because $f$ is analytic in $U$, I can choose two elements $z$ and $z_0$ of $U$ that satisfy $|z-z_0|=|\overline{z}-\overline{z_0}|<\delta$ and $|f(z)-f(z_0)|<\epsilon$. Since $|\overline{z}-\overline{z_0}|<\delta$, this means that $|f(\overline{z})-f(\overline{z_0})|<\epsilon$. Observing that $|f(\overline{z})-f(\overline{z_0})|=|\overline{f(\overline{z})-f(\overline{z_0})}|$, I conclude that $\overline{f(\overline{z})}$ is continuous.

This being the case, then (and I think I've done this part correct) I can compute the derivative by using the limit definition as:

$g'(z)=\lim\limits_{h\rightarrow{0}}\frac{\overline{f(\overline{z}+\overline{h})}-\overline{f(\overline{z})}}{h}=\lim\limits_{\overline{h}\rightarrow{0}}\overline{({f(\overline{z}+\overline{h})-f(\overline{z})})\ \ /\ \ {\overline{h}}}$, as I know that as $h\rightarrow{0}$, so does $\overline{h}\rightarrow{0}$. Since the function $f$ is analytic for all complex numbers in $U$, its derivative there, as given above, is the complex conjugate of the limit which is $\overline{f'(\overline{z})}$.

That's what I'm concerned about, however, because I haven't actually shown yet that $\overline{f(\overline{z})}$ is, in fact, analytic on its domain, all I know is that it is continuous on its domain. My thought is that it might only be analytic for the set $U$ if $U$ is a subset of the real numbers, since that is the only place that $\overline{z}$ is analytic and thus the composition $(a\circ{b})(z)$ for $b(z)=f(\overline{z})$ (analytic for $\overline{z}\in{U}$) and $a(z)=\overline{z}$, so the composition would be analytic on the set given by the intersection of the real numbers with $U$ but I don't believe that is right?

I've tried considering an equivalent case given by the conjugation $\overline{g(z)}=\overline{\overline{f(\overline{z})}}=f(\overline{z})$ to no avail.

Essentially, I'm asking to see if anyone can point me in the right direction and tell me if what I've done so far is on the right track.

I should add that these problems are from a section in the book preceeding the discussion about the Cauchy-Riemann Equations, so while I'm not sure if they will help, I don't want to use them.

I'm tagging this as homework although it's not 'homework' in the sense that I'm taking a class, but the problem is from a book, so I feel it is appropriate to tag it as such. I invite any admins to remove the tag if they feel it is unnecessary.

$$\lim_{h\to 0}\frac{\overline{f(\overline{z+h})}-\overline{f(\overline{z})}}{h}= \lim_{h\to 0}\frac{\overline{{f(\overline{z}+\overline{h})-f(\overline{z})}}}{h}=\overline{\lim_{h\to 0}\frac{{f(\overline{z}+\overline{h})-f(\overline{z})}}{\overline{h}}}=\overline{f'(\overline{z})}$$

• Yes, I have that. But that wasn't my question. May 25 '14 at 20:31
• This shows that $g(z)$ has derivative and that $g'(z)=\overline{f'(\overline{z})}$
– mfl
May 25 '14 at 20:34
• A complex function which has derivative is analytic. I have edited the typo involving the bar.
– mfl
May 25 '14 at 20:43
• I would agree that $f'(\overline{z})$ is analytic for $\overline{z}\in{U}$, but I don't see how it follows that $\overline{f'(\overline{z})}$ is analytic, much less $\overline{f(\overline{z})}$. Anyways, analytic functions have all their derivatives defined on the domain that they are analytic on, so what keeps the function in this case from having a second derivative that is undefined? May 25 '14 at 20:49
• We have that $g'(z)=\overline{f'(\overline{z})}.$ Repeating the same argument we get $g''(z)=\overline{f''(\overline{z})},$ and, by induction, $g^{(n)}(z)=\overline{f^{(n)}(\overline{z})}.$
– mfl
May 25 '14 at 21:49