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$f$ is analytic in {$z:|z|>1$} and $f(r)\in\mathbb R$ $\forall$ real $r>1$. How can I show that the same hold $\forall$ real $r<-1$?

Please don't solve it completely. I'm just looking for a clue. Thanks.

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    Hint: Consider the function $g(z) = \overline{f(\,\overline{z}\,)}$.2012-12-13
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    @LukasGeyer: Yup you're right.2012-12-13
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    @froggie: But $z\mapsto \overline z$ is not differentiable anywhere in $\mathbb C$.2012-12-14
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    It's true that $z\mapsto \overline{z}$ is not holomorphic; it is *antiholomorphic*, and thus $z\mapsto f(\,\overline{z}\,)$ is antiholomorphic. But if I take the conjugate again, $z\mapsto \overline{f(\,\overline{z}\,)}$ becomes holomorphic. If you're not comfortable with this argument, it is possible to check that $z\mapsto \overline{f(\,\overline{z}\,)}$ is holomorphic by appealing to the Cauchy-Riemann equations.2012-12-14
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    @froggie: Thanks.2012-12-14
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    @froggie You could post your comment as an answer (hence I will upvote).2013-04-22

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