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Exercise 3 of VII.4 of Conway's Complex Analysis states

Let $G$ be a simply connected region which is not the whole plane, and suppose that $\bar{z}\in G$ whenever $z\in G$. Let $a\in G\cap\mathbb{R}$ and suppose that $f:G\rightarrow D=\{z:|z|<1\}$ is a one-one analytic function with $f(a)=0,\ f'(a)>0$ and $f(G) = D$. Let $G_+=\{z\in G:\text{Im }z>0\}$. Show that $f(G_+)$ must lie entirely above or below the real axis.

I've been working on the proof via a suggestion given in another posting: show that $f(z) = \overline{f(\bar{z})}$ by looking at the conformal automorphisms of the unit disc. I see how the result follows directly if this claim is proven, but I'm having difficulty in showing the claim itself.

My work so far:

Let $g:D\rightarrow D$ be a conformal automorphism of the unit disc. Then $g$ must have the following form: $g(z) = e^{i\theta}\dfrac{z-\alpha}{1-\bar{z}\alpha}$.

I then look at $g(f(z))$ and demand that it have the same properties as $f$, namely, that $g(f(a)) = 0$ and $g'(f(a))f'(a)>0$. But then $g(z) = z$, which I then realized was obvious from the beginning, because $f$ is the unique function guaranteed by the Riemann Mapping Theorem....

I've also played around with specific curves in $G$. Let $C_1$ denote the real axis passing through $G$, and $C_2$ the perpendicular line passing through $a+iy$ and $a-iy$, where $y>0$ is such that $a+iy\in G$. Clearly, the angle between $C_1$ and $C_2$ is $\dfrac{\pi}{2}$ at $a$. Shifting $C_1$ by $it$, we maintain this right angle at $a+it$, $0\leq t\leq y$.

By the conformality of $f$ (and some unsightly calculations), I can then show that $f'(a+it) = f'(a-it)$, $0\leq t\leq y$. This is nifty. It'd be niftier if I could get it for all $a\in G\cap\mathbb{R}$ and then maybe do an integration trick.

Any help would be greatly appreciated. I am teaching myself these things in preparation for a qualifying exam, so nothing is too basic to point out.

(Meta comment: I would have asked this question in response to the cited posting, but my reputation isn't high enough yet. Also, I think the solution to this particular question must be simpler than the generalization provided in the previous posting.)

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    $\begingroup$ Let $h(z) = \overline{f\left(\overline{z}\right)}$. Once you see that $h$ is a biholomorphic map from $G$ to the unit disk with $h(a) = 0$ and $h'(a) = f'(a)$, look at $\alpha = f\circ h^{-1}$. That is an automorphism of the unit disk, with $\alpha(0) = 0$ and $\alpha'(0) = f'(a)\cdot (h^{-1})'(0) = f'(a)\frac{1}{h'(a)} = 1$. Knowing the Schwarz lemma, you can immediately deduce $\alpha = \operatorname{id}$ and hence $h = f$. $\endgroup$ – Daniel Fischer Aug 1 '14 at 18:07
  • $\begingroup$ And, actually, once I show that $h$ is biholomorphic and possesses the same properties of $f$ at $a$, I'd be done, since the Riemann Mapping Theorem guarantees that $f$ is the only function with those properties. $\endgroup$ – artificial_moonlet Aug 1 '14 at 18:49
  • $\begingroup$ True. Is there any problem in seeing that $h$ is biholomorphic? $\endgroup$ – Daniel Fischer Aug 1 '14 at 18:57
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Generally, for every holomorphic function $f \colon U \to \mathbb{C}$, the function $g \colon z \mapsto \overline{f\left(\overline{z}\right)}$ is holomorphic on the domain $V$ obtained by reflecting $U$ in the real axis. That is most easily seen by expanding $f$ into power series or the Wirtinger calculus. Both ways also show that we have $g'(z) = \overline{f'\left(\overline{z}\right)}$ on $V$.

Here, the domain $G$ and the image $\mathbb{D}$ are symmetric with respect to the real axis, so the complex conjugation is a bijection on each. Hence $h\colon z \mapsto \overline{f\left(\overline{z}\right)}$ is a bijection, since it is the composition of three bijections. By the first paragraph, it is also holomorphic, and thus biholomorphic. Further, we have $h'(a) = \overline{f'\left(\overline{a}\right)} = f'(a)$ since $a\in \mathbb{R}$ and $f'(a)\in \mathbb{R}$.

So $h$ and $f$ are biholomorphic maps between $G$ and the unit disk $\mathbb{D}$ with $h(a) = f(a) = 0$ and $h'(a) = f'(a) > 0$. With the uniqueness part of the Riemann mapping theorem it follows that $h = f$. Without that uniqueness part, the Schwarz lemma applied to the map $g = f\circ h^{-1}\colon \mathbb{D}\to \mathbb{D}$ which satisfies $g(0) = 0$ and $g'(0) = 1$ yields $g = \operatorname{id}_\mathbb{D}$, and hence $h = f$.

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  • $\begingroup$ How is it “most easily seen” depends on your preferred definition of a holomorphic function. For example, Cauchy–Riemann differential conditions are verified straightforwardly; Cauchy’s integral also doesn’t pose any problem. $\endgroup$ – Incnis Mrsi Nov 11 '14 at 20:06

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