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I was given this question : The only singularities in C (the complex set) of the analytic function $f$, are simple poles at z=1 and z=2, with residues at these poles equal to -3 and 7 respectively. If $f(0) = \frac{1}{2}$, determine the function f.

I have tried doing the following :

Using definition of residue at a simple pole and obtaining the following : $$\lim_{z\to2} f(z) = \frac{7}{z-2}$$

& $$\lim_{z\to1} f(z) = \frac{-3}{z-1}$$

But do not know how to go on from there. I have also tried using the Residue theorems by adding the given residues but did not manage either.

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    $\begingroup$ Actually such a function cannot be uniquely determined. For example, if $f(z)$ is a solution of this problem, then $g(z)=f(z)+z$ is another one. The general solution is $$f(z) = \frac{7}{z-2}-\frac{3}{z-1} +1 + zh(z)$$ where $h(z)$ is any entire function. $\endgroup$ – Crostul May 20 '16 at 7:23
  • $\begingroup$ As Crostul pointed out, the problem does not have a unique solution as stated. Read it carefully. Is there also an assumption about the behaviour at $\infty$? (E.g.. does it say the only singularities in $\overline{\mathbb{C}}$, the Riemann sphere?) $\endgroup$ – mrf May 20 '16 at 7:55
  • $\begingroup$ Thanks for your comments & help. It only says that the set C includes the point at infinity. The rest of the question is as above. $\endgroup$ – Andre Croucher May 20 '16 at 8:26
  • $\begingroup$ Thanks for your comments & help. It only says that the set C includes the point at infinity. The rest of the question is as above. $\endgroup$ – Andre Croucher May 20 '16 at 8:26

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