Let $$f(z) = \frac{z+1}{\sin ^2z}, a=0;$$ First we find the series for $\sin ^2z$: $$\sin^2 z = 1-\cos^2 z =\frac{1}{2}(1-\cos 2z)= \frac{1}{2}\left (1-\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(2z)^{2n}\right ) = \sum_{n=1}^\infty \frac{(-1)^n\cdot 4^n}{(2n)!}z^{2n}$$ which is valid in all of $\mathbb{C}$ because $\sin z$ is regular in $\mathbb{C}$ and the pointwise product (should be true for compositions, too) of regular functions is regular. So we have $$\frac{z}{\sin ^2z} + \frac{1}{\sin ^2z} \overset{?}= \frac{1}{\frac{\sin ^2z}{z}} +\frac{1}{\sin ^2z} = \frac{1}{\sum_{n=1}^\infty \frac{(-1)^n\cdot 4^n}{(2n)!}z^{2n-1}} + \frac{1}{\sum_{n=1}^\infty \frac{(-1)^n\cdot 4^n}{(2n)!}z^{2n}} $$
For the principal part I need something of the form $\sum_{k=-\infty}^{-1}\ldots$ and I seem to have made no progress whatsoever. I don't understand how I should get there.
The (bloody) textbook offers no examples about this problem, so I can't even tell if I'm on the right track.