# bounded real part then removable singularity

$f(z)$ 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 $f(z)$.

1) What do you know about the behaviour of $\psi(z)$ as $|z| \to \infty$?
• 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 $f(z)$ – ziang chen Feb 9 '13 at 14:32
• 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 $f$. – Robert Israel Feb 10 '13 at 9:13
• 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 $\varphi(z)$ – ziang chen Feb 10 '13 at 12:26