In the function $$f(z) =\frac{sin(\frac{\pi}{2}(z+1))}{(z^2+2z+4)(z+1)^3}$$ the order of the pole in $z=-1$ is correctly two? Or maybe it is an eliminable singularity?
I have a problem because often if I have an eliminable singularity I only have to Taylor expand the numerator of the fraction at the first term, in this situation, if I expand at the first term I have a singularity of order two and if expand at the second it disappears without becoming zero.
We have
$$\frac{\sin\frac\pi2(z+1)}{\left[3+(z+1)^2\right](z+1)^3}=\frac13\cdot\frac1{1+\left(\frac{z+1}{\sqrt3}\right)^2}\cdot\frac1{(z+1)^3}\sum_{n=0}^\infty\frac{(-1)^n\pi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$
$$=\frac13\left(1-\left(\frac{z+1}{\sqrt3}\right)^2+\left(\frac{z+1}{\sqrt3}\right)^4-\ldots\right)\frac1{(z+1)^{\color{red}2}}\sum_{n=0}^\infty\frac{(-1)^n\pi^{2n+1}(z+1)^{\color{red}{2n}}}{(2n+1)!}$$
and we can see clearly in the above Laurent expansion that the pole at $\;z=-1\;$ is of order two. Of course, the above is valid only in some neighbourhood of $\;-1\;$.