I need help with the following:
If $f$ is a function having $z_0$ as a second order Pole. and having the Laurent series $\sum_{k = - \infty}^{\infty} a_k (z-z_0)^k$. How could one calculate $Res_{z_0} f^2$ from the Laurent coefficients $a_k$ of $f$?
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MWSC1996
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What have you tried? – calc ll Jul 27 '22 at 00:45
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the laurent series of $f$ can be written as $ \frac{a_{-2}}{(z-z0)^2} + \frac{a_{-1}}{(z-z0)} + a_0 + \dots$ then I thought about doing the square and looking for the new $a_1$. The multiplication seems a bit difficult to me. – MWSC1996 Jul 27 '22 at 00:52
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You only need the new minus one term. – calc ll Jul 27 '22 at 00:59
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Write $f(z)=\sum_{k=-2}^{\infty}a_k(z-z_0)^k$.Multiply and get for the residue of $f^2$: $$c_{-1}=\sum_{i+j=-1}a_ia_j=2a_{-2}a_1+2a_{-1}a_0$$.
calc ll
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Thanks a lot! I just need to check where $(z-z_0)^-1$ remains after the multiplication. – MWSC1996 Jul 27 '22 at 01:17