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How to expand the complex function $$f(z)=\frac{1}{z^{2}+4}$$ at the region $|z-2i|>4$? Maybe i can use the standard geometric series with substitude?

Martin Ferrer suggested in comments that I use partial fractions. My partial fraction decomposition is $$f(z)=\frac{1}{4i}\left(\frac{1}{z-2i}-\frac{1}{z+2i}\right)$$ But now how do I use the fact that I am in the region $|z-2i|>4$?

Jacob Manaker
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Yanbo Liu
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Oct 09 '21 at 08:50
  • use $z^2+4=(z+2i)(z-2i)$ and partial fractions. Then you can use geometric series to both fractions. – Marcos Oct 09 '21 at 08:54
  • @MarcosEscartínFerrer Can you show me your steps,please? – Yanbo Liu Oct 09 '21 at 09:10
  • Yeah, but first I want you to do some work, you know how to do partial fractions? Where do you get stucked? Tell me what you tried and I can complete your answer. – Marcos Oct 09 '21 at 09:52
  • Yeah, here is my step: $\frac{1}{z^{2}+4}=\frac{1}{4i}(\frac{1}{z-2i}-\frac{1}{z+2i})$ , but since i want to expand the function at a region |z-2i|>4, how should i use the geometric series to achieve that? – Yanbo Liu Oct 09 '21 at 12:31
  • Notice that $\frac{1}{z+2i}=\frac{1}{z+4i-2i}=\frac{\frac{1}{4i}}{\frac{z-2i}{4i}+1}$, can you finish from this? – Marcos Oct 09 '21 at 18:04
  • @MarcosEscartínFerrer Ah, i see, thank you very much !! – Yanbo Liu Oct 10 '21 at 08:46
  • Maybe you can try to write a full solution, so If can check if you did it ok. Also It can be useful for more people. – Marcos Oct 10 '21 at 11:28
  • So after the arrangement, the equation will be f(z)=\frac{1}{4i}[\frac{1}{1-2i}-\frac{1}{z+2i}] – Yanbo Liu Oct 11 '21 at 15:36
  • @MarcosEscartínFerrer But just a quick question, what should I do about the term \frac{1}{z-2i}, how should i expand this term at the region |z-2i|>4???? – Yanbo Liu Oct 11 '21 at 15:43
  • You are expanding in powers of $z-2i$ so this is already expanded. – Marcos Oct 11 '21 at 15:46
  • omg, what was i think. Thank you , now i totally understand it. – Yanbo Liu Oct 11 '21 at 15:50

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f(z)=$\frac{1}{z^{2}+4}$=$\frac{1}{4i}(\frac{1}{z-2i}-\frac{1}{z+2i})=\frac{1}{4i}(\frac{1}{z-2i}-\frac{\frac{1}{4i}}{1+\frac{z-2i}{4i}})=\frac{1}{4i}(\frac{1}{z-2i}-\frac{1}{4i}\sum_{n=0}^{n=\infty}\frac{(-1)^{n}(4i)^{n}}{(z-2i)^n})$ valid when $|\frac{z-2i}{4i}|>1$ which is $|z-2i|>4$

Yanbo Liu
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