I am tasked with finding whether the function $()=\tan(/(-1))$ can be developed into a Laurent series around $=1$, and if so what is the radius of convergence and what is the residue?
My initial attempt was at developing the expression into something like the following:
$ \tan(\frac{z}{z-1}) = \tan(1 + \frac{1}{z-1}) = \frac{\sin(1 + \frac{1}{z-1})}{\cos(1 + \frac{1}{z-1})} = \frac{ \sin(1)\cos(\frac{1}{z-1}) - \cos(1)\sin(\frac{1}{z-1}) }{ \cos(1)\cos(\frac{z}{z-1}) - \sin(1)\sin(\frac{z}{z-1}) } $
I know how to develop either the numerator or denominator into taylor series expansions based on this, but seems to me like would get quite hairy to explicitly try to develop the quotient (possible but I don't believe thats the purpose of the exercise).
A couple of other directions:
try to show that $f(z)$ is indeed defined and analytic in each punctured disc of $z=1$ but that there are two sequences of complex numbers converging to 1 that have different limits and therefore it is an essential singularity (e.g. $\lim_{n->\infty}f(a_n) \ne \lim_{n->\infty}f(b_n)$). perhaps something like $a_n = 1 + \frac{1}{2n\pi - 1}$ and $b_n = 1 + \frac{1}{2(n+1)\pi - 1}$
try to show that at $z = 1$ we do not actually have an isolated singularity since we can find a point at any neighborhood of $z=1$ in which $f(z)$ is not defined? for example at $a_n = 1 + \frac{1}{(2n+1)\frac{\pi}{2} - 1}$ for every n the numerator is zero and therefore it is undefined, and $\lim_{n->\infty}a_n = 1$ I believe?