Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

Hints:

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

2) Do you know the Casorati-Weierstrass theorem?

share|improve this answer
    
Please in detail. Thanks a lot –  ziang chen Feb 8 '13 at 9:19
    
What have you done so far? Where are you stuck? –  Robert Israel Feb 8 '13 at 18:54
    
Thanks! ok! $\mathrm{R}'>\mathrm{R}$ –  ziang chen Feb 9 '13 at 14:08
    
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.