I am working through problems in "Cracking the GRE Mathematics Subject Test" (4th edition) and one of the practice problems asks for the coefficient of $(z - 2)^{-1}$ in the Laurent series for $f(z) = \frac{1}{z-5}$ centered at $z = 2$. For my solution I did $$\frac{1}{z-5} = -\frac{1}{3 - (z - 2)} = -\frac{1}{3}\frac{1}{1 - \frac{z - 2}{3}} = \sum_{n = 0}^\infty{3^{-n - 1}(z - 2)^n}.$$ This expansion is valid on $|z - 2| < 3$. Hence, the coefficent is $0$. However, for the solution in the book they say $$\frac{1}{z-5} = \frac{\frac{1}{z - 2}}{1 - \frac{3}{z - 2}} = \frac{1}{z-2}\sum_{n = 0}^\infty{\left(\frac{3}{z-2}\right)^n} = \sum_{n = 0}^\infty{3^n(z-2)^{-n - 1}}.$$ This expansion is valid on $|z - 2| > 3$. Hence, the coefficent, using this expansion, is 1. Is there a reason that the second one is correct and the first one is not?
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1$\begingroup$ Observe that $\;3-(z-2)=z+5\;$ ...and that is not what you have in the very first expression's denominator. $\endgroup$– DonAntonioSep 4, 2022 at 19:51
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$\begingroup$ In the denominator in the first expression I have $-(3-(z-2)) = -(3 - z + 2) = -(5 - z) = z - 5$ which is indeed the same as the original denominator. $\endgroup$– PaulSep 4, 2022 at 20:55
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2$\begingroup$ Now that's right with the minus sign in the second expression. And the reasong why they did that' apparently, is because in our first expression you got a Taylor series, of if you prefer: a Laurent series with principal part euqal to zero, whereas in the second expression you did actually get a Laurent series with non-zero principal series $\endgroup$– DonAntonioSep 4, 2022 at 21:45
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$\begingroup$ I think they are both Laurent series which are just valid on different domains. The first one is also a Taylor series, but that's because it is analytic at 2. But I think that you're probably right that they chose the second one because you get a Laurent series which actually has negative exponent terms. $\endgroup$– PaulSep 8, 2022 at 0:14
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$\begingroup$ Hope this helps: math.stackexchange.com/questions/1921924/… $\endgroup$– Hussain-AlqatariSep 10, 2022 at 9:20
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