I know that Laurent's series is used for an annulus with the function $f(z)$ being analytic in the region R defined by the annulus ($r<|z|<R$, where $R > r$).
Now my question is that Laurent's series has both positive and negative coefficients, so do the negative coefficients (along with the whole term) represent the non-analytic portions (or the singularities contained) inside $|z|<R$ which is essentially the region defined by $|z|\leq r$ ?
Also if we make the region $|z|\leq r$ analytic for the function $f(z)$, does that mean we have converted the annulus to a disc and the Laurent's series to Taylor's as that would cause the negative coefficients to become zero?
Lastly if the annulus is there (i.e. we have not removed the singularities), does that mean the positive coefficients correspond to Taylor's coefficient for that region R ?
Edit: By saying positive and negative coefficients I mean $a_{n}$ and $a_{-n}$ respectively which essentially corresponds to positive and negative exponents.
The way that you have explained your question gives the impression that we are in control of the regions of convergence of series and we can change the disk to annulus and or vice versa as we wish.
Well, it is the function which dictates the region of convergence and we do not change it.
If the function does not have any singularities, then you do not have any negative exponents and the region where Taylor Series converges, normally is in form of $$|z-a|\le R$$ On the other have if there are poles or essential sigularities, we have negative exponents and we get two types of restrictions for the region the series converges which is normally in form $$ r<|z-a|<R$$.