Suppose $f$ is analytic in the semi-disc: $|z| \le 1$, $\operatorname{Im}(z) > 0$ and real on the semi-circle $|z| = 1, \operatorname{Im}(z) > 0$. Show that if we set
$$
g(z) =
\begin{cases}
f(z) & \text{if $|z| \le 1, \operatorname{Im}(z) > 0$} \\ \\
\overline{f\left(\frac{1}{\overline z}\right)} & \text{if $|z| > 1, \operatorname{Im}(z) > 0$} \\
\end{cases}
$$
then $g$ is analytic in the upper half-plane.
Should I use specific theorems?I know about schwartz reflection principle. I wonder if it has a straightforward solution.
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I mean... if you know Schwartz Reflection principle then apply it... I don't get the question: have you tried and failed? Because before asking it should be best to try – b00n heT May 16 '19 at 17:26
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@b00nheT I have no idea how to apply the theorem. – FreeMind May 16 '19 at 17:28
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1You forgot to conjugate the $z$ in the denominator $\overline{f(1/\bar{z})}$. This is the reflection principle applied to circular arcs instead of intervals, namely reflect, apply $f$, and reflect again. – user10354138 May 16 '19 at 17:28
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@user10354138 Oh, you are right. I corrected the question – FreeMind May 16 '19 at 17:29
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You can either deduce from Morera's theorem, or you can find some Moebius transformation mapping the circle to the real line and apply Schwartz's reflection principle. – logarithm May 16 '19 at 17:46