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I search the Laurent series of $\sin(\frac{z}{z+1})$ in the point $z_0=-1$. What I did is write the taylor expansion

$$\sin\Big(\frac{z}{z+1}\Big)=\sum_{n=0}^\infty(-1)^n\frac{\Big(\frac{z}{z+1}\Big)^{2n+1}}{(2n+1)!}$$

But this can't be right, since I didn't take $z_0$ into account. How can I do the Laurent expansion in this case?

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$$\frac z{z+1}=1-\frac1{z+1}$$ so $$\sin\left(\frac z{z+1}\right)=\sin\left(1-\frac1{z+1}\right) =\sin 1\cos\left(\frac1{z+1}\right)-\cos1\sin\left(\frac1{z+1}\right).$$ Can you take it from here?

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  • $\begingroup$ $\sin 1\cos\left(\frac1{z+1}\right)-\cos1\sin\left(\frac1{z+1}\right)= \\ \sin1\sum_{n=0}^\infty(-1)^n\frac{1}{(z+1)^{2n}(2n)!} - \cos 1\sum_{n=0}^\infty(-1)^n\frac{1}{(z+1)^{2n+1}(2n+1)!}\\$ I would get the same result for $z_0=2$. The expansion is the same in each point then? $\endgroup$ – Kiji Apr 14 '18 at 14:54
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A solution using the "hard way". For some $f\in C^\omega(\Bbb C\setminus\{z_0\},\Bbb C)$ the Laurent expansion around $z_0$ is defined by

$$f(z)=\sum_{n=1}^\infty c_{-n}(z-z_0)^{-n}+\sum_{n=0}^\infty c_n(z-z_0)^n\tag1$$

And it is known that

$$c_n=\frac1{2\pi i}\oint_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^{n+1}}\, dz,\quad n\in\Bbb Z,\, r>0\tag2$$

Also we have the Cauchy derivative formula

$$f^{(k)}(z_0)=\frac{k!}{2\pi i}\oint_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^{k+1}}\, dz,\quad k\in\Bbb N_{\ge 0},\,r>0\tag3$$

Thus for the Laurent expansion of $g(z):=\sin\left(\frac{z}{z+1}\right)$ around $z_0=-1$ we have that

$$\begin{align}c_n&=\frac1{2\pi i}\oint_{|z+1|=1}\frac{g(z)}{(z+1)^{n+1}}\, dz\\ &=\frac1{2\pi i}\oint_{|z+1|=1}\sum_{k=0}^\infty\frac{(-1)^k z^{2k+1}}{(2k+1)!(z+1)^{2k+n+2}}\, dz\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{2\pi i(2k+1)!}\oint_{|z+1|=1}\frac{ z^{2k+1}}{(z+1)^{2k+n+2}}\, dz\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+n+1)!(2k+1)!}\partial^{2k+n+1}[z^{2k+1}]_{z=-1}[2k+1\ge 2k+n+1\ge 0]\\ &=(-1)^n\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\cdot\frac{(2k+1)^\underline{2k+n+1}}{(2k+n+1)!}[2k+1\ge -n\ge 0]\\ &=\frac{(-1)^n}{(-n)!}\sum_{k=0}^\infty\frac{(-1)^k}{(2k+n+1)!}[2k+1\ge-n\ge 0]\\ &=\displaystyle\begin{cases}\frac{(-1)^n}{(-n)!}\sin (1),&-n\ge 0\text{ and }n\text{ is even}\\\frac{(-1)^n}{(-n)!}\cos (1),&-n\ge0\text{ and }n\text{ is odd}\\0,&-n<0\end{cases}\end{align}$$

where $[\cdots]$ is an Iverson bracket, and $a^\underline b$ is a falling factorial. Thus the auxiliary part of the Laurent series of $g$ around $-1$, but $c_0$, is zero.

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  • $\begingroup$ Very clear derviation, thanks. The solution in the comment I made above is wrong then, since there are $\sin 1$ and $\cos 1$? $\endgroup$ – Kiji Apr 14 '18 at 17:38
  • $\begingroup$ @Kiji Im not sure if my solution is correct (it seems), the answer of Lord seems very clear, and it imply that the coefficients are of the form $(-1)^k\frac{\sin 1}{2k}$ and $(-1)^{k+1}\frac{\cos 1}{2k+1}$ $\endgroup$ – Masacroso Apr 14 '18 at 17:40
  • $\begingroup$ Yeah, thank to you both (I dont have enough reputation yet to upvote) $\endgroup$ – Kiji Apr 14 '18 at 17:46
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    $\begingroup$ @Kiji I had some mistake before that I corrected now. My answer coincide with that of LordShark. $\endgroup$ – Masacroso Apr 14 '18 at 18:48

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