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Background Wunsch-Complex Variables 3e 5.6#6

My attempt: $\sum_{0}^\infty\frac{1}{z}\left (\frac{-i}{z}\right )^n=(\frac{1}{z}-\frac{i}{z^2}-\frac{1}{z^3}+\frac{i}{z^4}\dots)$

Author's solution: $\sum_{-\infty}^{-1}(\color{red}{-1})\left ((-i)^{1-n}z^n\right )=(\frac{1}{z}-\frac{i}{z^2}+\frac{i^2}{z^3}+\frac{\color{red}{i^3}}{z^4}\dots)$

Did I make a mistake? Where did red -1 come from?

I can show my steps but it's just a straight forward application of the formula you can find here as described by Markus Scheuer.

Community
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user5389726598465
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2 Answers2

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Your formula is fine, so is the authors. They just use a different enumeration. The two ways are related by $n\mapsto -1-n$. Starting with $1-(-1)=2$ they add the factor $(-1)=i^2$ in front. There is indeed a sign misprint regarding their coefficient to $z^{-4}$. I think yours is more clear.

H. H. Rugh
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  • Yes, his formula is right, it was a typo on $c_4$. Funny that the first time I read it I mentally calculated the same value as the typo. – user5389726598465 Sep 02 '16 at 13:11
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First put $$\omega=\frac{1}{z}$$ Function becomes $$\frac{\omega}{\omega+i}\\$$ Now expanding$ \frac{1}{\omega+i} $by binomial theorem we get $$\frac{1}{\omega+i}=1-i\omega+({i\omega})^2-......\\$$ now put this result in modified function and replace $\omega=\frac{1}{z}$

user2104628
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