the Laurent series of $f(z)=\cos \frac {z}{z+1}$ on $0<|z-1|<\infty$

Question

I want to find the Laurent series of $f(z)=\cos \frac {z}{z+1}$ on $0<|z-1|<\infty$, and I don't understand, if it is ok, that a point $-1$ is in $0<|z-1|<\infty$, also I don't understand how to find the Laurent series. I'm looking for any hints. Thank you in advance!

Answer 1

Use the fact that\begin{align}\cos\left(\frac z{z+1}\right)&=\cos\left(1-\frac1{z+1}\right)\\&=\cos(1)\cos\left(\frac1{z+1}\right)+\sin(1)\sin\left(\frac1{z+1}\right)\\&=\cos(1)\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!(z+1)^{2n}}+\sin(1)\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!(z+1)^{2n+1}}\\&=\cos(1)+\frac{\sin(1)}{z+1}-\frac{\cos(1)}{2(z+1)^2}-\frac{\sin(1)}{6(z+1)^2}+\cdots\end{align}