Finding The Laurent Series for $\sin(\frac{z}{1+z})$ at $z = -1.$

Question

I'm trying to find the Laurent Series for $\sin\left(\frac{z}{1+z}\right)$ centered around $z = -1.$ It seems to me that I should be considering the typical power series expansion of $\sin$, i.e.

$$\sin(w) = \sum_{k = 0}^{\infty} (-1)^k \frac{w^{2k+1}}{(2k+1)!}.$$

However, when I perform this with our current function, I see that

$$\sin\left(\frac{z}{1+z}\right) = \sum_{k = 0}^{\infty} (-1)^k \frac{z^{2k+1}}{(1+z)^{2k+1}(2k+1)!}.$$

I am not entirely sure what to do with $z^{k+1}.$ Are there some other things that I should be doing with this series to account for the singularity of $\frac{z}{1+z}$ at $z = -1$?

Answer 1

HINT:

$$\sin\left(\frac{z}{z+1}\right)=\sin\left(1-\frac{1}{z+1}\right)=\sin(1)\cos\left(\frac{1}{z+1}\right)-\cos(1)\sin\left(\frac{1}{z+1}\right)$$

Answer 2

$$ \begin{align} \sin\left(1-\frac1{1+z}\right) &=\sin(1)\cos\left(\frac1{z+1}\right)-\cos(1)\sin\left(\frac1{z+1}\right)\\ &=\sin(1)\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\left(\frac1{z+1}\right)^{2k}-\cos(1)\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\left(\frac1{z+1}\right)^{2k+1} \end{align} $$