I was solving this problem: find the two Laurent Series representations of $f: \mathbb{C}\setminus\{0,i,-i\}\to\mathbb{C}$ given by $f(z) = 1/z(z^2+1)$ in the correct domains.
My approach was: I've used partial fractions to rewrite $f$ as
$$f(z)=\dfrac{1}{z}-\dfrac{1}{2}\dfrac{1}{z-i}-\dfrac{1}{2}\dfrac{1}{z+i},$$
then I've done as follows, if $0<|z|<1$, then both $z \mapsto 1/(z-i)$ and $z\mapsto 1/(z+i)$ are analytic and can be expressed in their Taylor Series. For that matter, I rewrite
$$-\dfrac{1}{2}\dfrac{1}{z-i}=\dfrac{1}{2i}\dfrac{1}{1-z/i}=\dfrac{1}{2i}\sum_{n=0}^\infty \dfrac{z^n}{i^n}=\dfrac{1}{2}\sum_{n=0}^{\infty}\dfrac{z^n} {i^{n+1}},$$
and also
$$-\dfrac{1}{2}\dfrac{1}{z+i}=\dfrac{1}{2}\dfrac{1}{-z-i}=\dfrac{1}{2i}\dfrac{1}{-1-z/i}=\dfrac{i}{2}\dfrac{1}{1-(iz)}=\dfrac{i}{2}\sum_{n=0}^\infty i^n z^n=\dfrac{1}{2}\sum_{n=0}^\infty i^{n+1}z^n,$$
thus we should have
$$f(z) = \dfrac{1}{z} + \dfrac{1}{2}\sum_{n=0}^{\infty}\dfrac{z^n} {i^{n+1}}+\dfrac{1}{2}\sum_{n=0}^\infty i^{n+1}z^n,$$
however the answer is $1/z+ \sum_{n=0}^{\infty}(-1)^{n+1}z^{2n+1}$. Now, what I've done wrong there? Is there something wrong I did, or there's just some tricky to simplify this last expression I didn't see?
Thanks very much in advance!
Why won't you expand directly?! If you have two series you will have to check very carefully whether some summands of one of them add/cancel with some summands of the other one (and this is what happens here...).
$$|z|<1\implies\frac1z\frac1{1+z^2}=\frac1z\sum_{n=0}^\infty(-1)^nz^{2n}$$
based on the powers expansion
$$\frac1{1+z}=\sum_{n=0}^\infty (-1)^nz^n\;,\;\;|z|<1$$