Approssimazione in serie di Laurent

[Boomerang]

Buona sera, ho problemi con il seguenete esercizio:
-determinare lo sviluppo in serie di Laurent delle seguenti funzioni relativamente ai punti indicati:
$f(z)=(z-3)sen (1/(2+z)) , z_0=-2$. Non capisco perche' nella soluzione l'approssimazione nel punto indicato e'data da: $f(u(z))=(u-5)*(1/u-1/(3!u^3)+1/(5!u^5)+...)$, avendo applicato la sostituzione $u(z)=z+2$. Quella presentata dovrebbe essere la soluzione per $z=oo$ o forse mi sto confondendo? Grazie.

[javicemarpe]

The Laurent series of $sin z$ at $infty$ is that one you get, but your function is not $sin z$ but $sin(1/z)$ (up to that change of variables you do).

[Boomerang]

So is the expansion wrong or not?

[javicemarpe]

No, it's not wrong. As I told you, the series you get would be the Laurent expansion of $sin z$ at $infty$, but your function is $sin \frac{1}{z}$, so its expansion at $0$ is the same as the expansion at $\infty$ of $sin z$ because of definition of Laurent series expansion of a function at $\infty$.

Summarising: your result is correct and your confusion comes from the definition of Laurent expansion of a function $f(z)$ at $\infty$ (which is the Laurent expansion of $f(\frac{1}{z})$ at $0$) and the fact that your function is $sin \frac{1}{z}$ instead of $sin z$.

[Boomerang]

Ok... so this expansion of $f(z)=sen(1/z)$ is true for all point in $ CC $ with the exception of $z_0=0$; therefore ,until we exlude this point from the circular crown (in wich f is olomorphic), the approximation is valid (thanks to the laurent's expansion theorem) and suggest to us what is his comportament nearby $z_0=0$, in analogy with $sen (z) $ in an infinite point. Is it right?

[javicemarpe]

Yes, it is.

[Boomerang]

Thank you for the answer,javi.

[javicemarpe]

Prego.

[Boomerang]

another two questions:
1) if i have a $z_0$ point in wich $f(z_0)$ has got a n order pole (for exemple $n=1$) so the Laurent's series will be expressed ever like $f(z)=c_(-1)/(z-z_0)+sum_(n=0)^(oo)a_n(z-z_0)^n$?

2) What the differences between study a function in a point and in a intervall?
Take for exemple $f(z)=1/[(z+1)(z+3)]$ and the $A:={1<|z-0|<3}$. It's easy to see that the function has not problems in $A$ beacuse we exclude the singular points and the expansions are definited in $z_0=0$ that is not a singular point (like we can see calculating $f(z_0)$. If we decompose the $f(z)=f_1(z)+f_2(z)=A/(z+1)+B/(z+3)$ why in one case (for $|z|>1$) we do: $f_1(z)=A/[z(1+1/z)]$ while in the second (for $|z|<3$): $f_2(z)=B/[3(1+z/3)]$? And how much important is decompose in fraction? Is there any method more directly than the decomposition? Thank you.

[javicemarpe]

1) Yes.

2) I don't understand your question. I mean, "decompose in (partial) fractions" for what?