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I need to find the principal part of the Laurent expansion of the function in the annulus described in the title.

In my attempt, I have replaced $z$ by $w:=z - \sqrt[3]{2 \pi}$. I know the Taylor expansion of the $\sin$ function. I have tried to use a approach similar to the one used in this answer, where we know $g(z)$ and try to compute coefficients one by one of the function $f(z):=1/g(z)$ using the fact that $f(z)g(z)=1$.

One of the doubts I am having with this method is: why can we write such an $f$ without negative powers of $z$? If I can do that in our example then I guess all of the negative coefficients are equal to $0$... Can you please help me clarify and give some hints to the needed computation? Thanks a lot.

Barreto
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    If $g(z):=(z-a)f(z)$ is holomorphic in a neighbourhood of $a$ the principal part consists of just one term $g(a)/(z-a).$ The value $g(a)$ can be calculated as the limit $\lim_{z\to a}(z-a)f(z)$ – Ryszard Szwarc Mar 25 '22 at 21:31
  • Thank you for your answer @RyszardSzwarc. In your notation, $f(z)=\frac{1}{\sin z^3}$ and $a=\sqrt[3]{2 \pi}$ right? I will try to use your hint although computing the complex limit does not seem an easy task.. – Barreto Mar 25 '22 at 21:41
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    Exactly. Calculate the reciprocal of the limit and notice that it has the form of the derivative of the function $\sin(z^3)$ at $a.$ – Ryszard Szwarc Mar 25 '22 at 21:45
  • Is the final answer $c_{-3}=1/3$ and $0$ for all other coefficients corresponding to negative powers? I am trying some other approach now..@RyszardSzwarc – Barreto Mar 25 '22 at 23:11
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    No. I got $c_{-1}={\sqrt[3]{2\pi}\over 6\pi}.$ It seems that you calculated the Laurent expansion at $0,$ not at $ \sqrt[3]{2\pi}.$ By the way, at $0$ the result would be $c_{-3}=1.$ – Ryszard Szwarc Mar 26 '22 at 00:30
  • Dear @RyszardSzwarc I finally got some answer that makes sense. I agree that there is only one non zero coefficient in the principal part, but I go that $c_{-1}=3(\sqrt[3]{2 \pi})^2$. Are you sure of your answer? – Barreto Mar 27 '22 at 15:32
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    I got the reciprocal $${1\over 3(\sqrt[3]{2\pi})^2}= {\sqrt[3]{2\pi}\over 6\pi}$$ – Ryszard Szwarc Mar 27 '22 at 15:58
  • My apologies, I meant that!. Did it. Thank you! – Barreto Mar 27 '22 at 16:03

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