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**Bruno J.** $\displaystyle \frac{1}{2z-i}=\frac{1}{2(z+1)-(2+i)} = \frac{-1}{2+i}\frac{1}{1-\frac{2}{2+i}(z+1)} = \frac{-1}{2+i}\sum_{j=0}^\infty \left(\frac{2}{2+i}(z+1)\right)^j$ for $\displaystyle \left|\frac{2}{2+i}(z+1)\right|<1$.

I don't know how you got logs in there!

The Taylor series of a function around a point of analyticity *is* a specific case of a Laurent series : the Taylor series is the Laurent series there. However, around a singularity, there is no Taylor series, but there may be a Laurent series. In this case, since the function is analytic at $\displaystyle -1$, the Laurent series you are being asked to find is nothing else than the Taylor series there.