Find the principal part for the following Laurent series: $$\frac{z^{2}}{z^{4}-1}$$ for $0< |z-i| <\sqrt{2}$
Attempt:
$$\frac{z^{2}}{z^{4}-1} = \frac{1}{2(z^2+1)} + \frac{1}{2(z^2-1)} = \frac{1}{2(z+i)(z-i)} + \frac{1}{2(z-1)(z+1)} \text{ (1)}$$
Also, $$\frac{1}{z+1} = \frac{1}{(1+i)(1 + \frac{z-i}{1+i})} = \frac{1}{1+i}\sum_{n=0}^{\infty}(-1)^n(\frac{z-i}{1+i})^n$$
and
$$\frac{1}{z+i} = \frac{1}{2i(1 + \frac{z - i}{2i})} = \frac{1}{2i}\sum_{n=0}^{\infty}(-1)^n(\frac{z-i}{2i})^n$$
Therefore, principal part from LHS in (1) is $\frac{1}{4i(z-i)}$ and from RHS, it will be $0$ I guess. Therefore, principal part should be $\frac{1}{4i(z-i)}$.
Am I correct? I am still not sure about the last part (principal part of $\frac{1}{2(z-1)(z+1)}$ being $0$). Can anyone give tips?
Observe that in the given domain your function only has one simple pole, namely $\;z=i\;$, and thus the principal part of its Laurent series there will contain only one single element, namely $\;\cfrac{a_{-1}}{z-i}\;$ . In this case, thus, the easiest way to solve your problem is evaluating the residue $\;a_{-1}\; $ using Cauchy's Formula:
$$a_{-1}=\lim_{z\to i}(z-i)f(z) =\lim_{z\to i}\frac{z^2}{(z+i)(z^2-1)}=\frac{-1}{2i(-1-1)}=\frac1{4i}$$
and thus yes: your work is correct.