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1) A holomorphic function $f$ on the disc $\{z \in \mathbb{C}: |z|< 2\}$ such that $f(1/n) = (-1)^n/n$ for every positive integer $n$.

2) A rational function $f$ having a pole at 0 such that the residue of $f$ at 0 equals 2 and the residue of the derivative $f'$ at 0 equals 1.

A group of us have plugged away at these for a while, but have no idea what we're missing.

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1) Try f(n)=ncos(n). This is a product of two holomorphic functions and is thus holomorphic with f(1/n) = .. as desired. –  Arbias Hashani Dec 12 '12 at 1:50
    
@ArbiasHashani: if you define $f$ in that way, $f(1/n) = (1/n)\cos(1/n)$. But $\cos(1/n)\neq (-1)^n$.... –  froggie Dec 12 '12 at 1:55
    
ah yes, apologies! i was thinking of values of Pi instead. thanks for correcting me froggie. –  Arbias Hashani Dec 12 '12 at 1:58

1 Answer 1

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Hints:

For (1), consider the function $g = f^2$, so that $g(1/n) = (1/n)^2$ for every positive integer $n$. Using the Identity Theorem, what can you say about $g$? Using this, what must $f$ be?

For (2), the assumption is that $f$ looks like $\cdots + \frac{2}{z} + a_0 + a_1z + \cdots$. Compute the derivative of this term by term. This gives you $f'$. What is the residue of $f'$?

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