Let $V(z)$ vector space spanned by $e(k),o(k)$. We define them as follows.
$$e^k(z):=\Bigg(\frac{\partial}{\partial z}\Big(z\frac{\partial}{\partial z}\Big)^{2k}\frac{z^2}{(1-z^2)}+\delta_{k,0}\frac1z\Bigg)dz $$
$$o^k(z):=\Bigg(\frac{\partial}{\partial z}\Big(z\frac{\partial}{\partial z}\Big)^{2k}\frac{z}{(1-z^2)}\Bigg)dz $$
So $$e(0)= \frac{\partial}{\partial z}\frac{z^2}{1-z^2}+\frac1z$$
We observe that $\forall k\geq 0$ by computation $$Res_{z=0} e(k)=-Res_{z=1} ln(z)\ e(k)-Res_{z=-1} ln(z)\ e(k)$$ where $Res$ denote the residue at the point. $ln(z)$ denote the log function. This is also true for $o(k)$. That is $$Res_{z=0} o(k)=-Res_{z=1} ln(z)\ o(k)-Res_{z=-1} ln(z)\ o(k)$$
Can anyone give the proof of the above fact?
I made an observation that $$o^{k+1}= z^2 \frac{\partial^2}{\partial z^2}o^{k}(z)+2z\frac{\partial}{\partial z}o^k (z) +z\frac{\partial}{\partial z}o^k (z) +o^{k}(z) $$ For $k>0$ we can see that $$Res_{z=0}o^k (z)=0 , Res(ln(z)z^2 \frac{\partial^2}{\partial z^2}o^{k}(z),z=1)=0, Res(ln(z)2z\frac{\partial}{\partial z}o^k (z),z=1)=0,Res(ln(z)z\frac{\partial}{\partial z}o^k (z),z=1)=0, Res(lno^{k}(z),z=1)=0 $$ Then by induction can we conclude our proof?