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$f(z)$ is analytic on $R<|z|<+\infty$, and $|\mathrm{Re}f(z)|\leq M$. The Laurent series Expansion $$f(z)=\varphi(z)+\psi(z),$$ where $\varphi(z)=\sum\limits_{n=0}^\infty a_nz^n$ is the principal part of $f(z)$ at $\infty$. Show that $\mathrm{Re} \varphi(z)$ is bounded, and then prove $\varphi(z)$ is constant, so $\infty$ is a removable singularity of $f(z)$.

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Hints:

1) What do you know about the behaviour of $\psi(z)$ as $|z| \to \infty$?

2) Do you know the Casorati-Weierstrass theorem?

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  • $\begingroup$ Please in detail. Thanks a lot $\endgroup$ – ziang chen Feb 8 '13 at 9:19
  • $\begingroup$ What have you done so far? Where are you stuck? $\endgroup$ – Robert Israel Feb 8 '13 at 18:54
  • $\begingroup$ Thanks! ok! $R'>\mathrm{R}$, $\psi(z)$ is bounded on $ R'\leq|z|<+\infty$, so is $\mathrm{Re}\psi(z)$.then $\mathrm{Re}\varphi(z)$ is bounded on $ R'\leq|z|<+\infty$, too. $\varphi(z)$ is an entire function, so $\varphi(z)$ is bounded on $ |z|\leq R'$. we get that $\mathrm{Re}\varphi(z)$ is bounded on $ \Bbb C$. then,we have two ways.1).According to Liouville theorem, $ \varphi(z)$ is constant. Or 2) According to Casorati-Weierstrass theorem, $\infty$ is not an essential singularity of $f(z)$; According to Fundamental Theorem of Algebra,$\infty$ is not a pole of $f(z)$ $\endgroup$ – ziang chen Feb 9 '13 at 14:32
  • $\begingroup$ 1) The Liouville theorem would work if you knew $\varphi$ was bounded, but you only know about $\text{Re}(\varphi)$. 2) You want to apply Casorati-Weierstrass and Fundamental Theorem of Algebra to $\varphi$, not to $f$. $\endgroup$ – Robert Israel Feb 10 '13 at 9:13
  • $\begingroup$ Thanks! en. $\mathrm{Re}\varphi(z)$ was bounded, it was easy to prove $\varphi(z)$ was constant with Liouville theorem. every textbook contain this proof; 2) yes, I apply Weierstrass and Fundamental Theorem of Algebra to $\varphi(z)$ $\endgroup$ – ziang chen Feb 10 '13 at 12:26

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