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.

$f:D\rightarrow\mathbb{C}$ be analytic satisfying $f(1/n)=\frac{2n}{3n+1}$, $D$ is open unit disk,Then

1.$f(0)=2/3$

2.$f$ has a simple pole at $z=-3$

3.$f(3)=-3$

4.No such $f$ exist.

well, $1$ is correct due to continuity of $f$, define $g(z)=f(z)-\frac{2}{3+z}$ then $g(1/n)=0$ and $0$ is a limit point of set of zeroes of $g$ and hence $f(z)=\frac{2}{3+z}$ hence $2$ is also correct.am I right?

share|improve this question
    
Are $3$ and $-3$ inside your open unit disk? ;) –  N. S. Jun 13 '12 at 15:54
    
:P :P :P :P :P :P extremely sorry.thnx for correcting :) –  Une Femme Douce Jun 13 '12 at 15:58
    
Ok so only $1$ is correct. –  Une Femme Douce Jun 13 '12 at 16:00
1  
Yup. What is true about (2) is that $f$ has an analityc continuation to $\mathbb{C}\backslash \{1 \}$, and this analytic continuation satisfies (2). But this is a new function... –  N. S. Jun 13 '12 at 16:07

1 Answer 1

up vote 2 down vote accepted

What you did is correct. We have to consider the analytic extension to $\Bbb C\setminus \{-3\}$ after having proved it exists. So 1. and 2. are correct.

share|improve this answer

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.