I want to find the Laurent series of $1\over(z^2+i)$at $z=i$. How do I start and go on from there? Do I start with either of these?
$$1 \over{(z+i\sqrt i)(z-i\sqrt i)}$$
$$\frac{1}{i} \frac{1}{1+(z^2/i)}$$
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1First do partial fractions. If we abbreviate $w=(1-i\sqrt2)/2$, then the zeros of $z^2+i$ are $w$ and $-w$. Then you can find constants $A$ and $B$ such that $$\frac1{z^2+i}=\frac A{z-w}+\frac B{z+w}.$$ Then you can write the denominators in the form $C+D(z-i)$ (different values of $C,D$ for the two terms), and manipulate that to a form where you can use the geometric series sum formula. – Jyrki Lahtonen Dec 07 '18 at 04:55
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given answer is https://i.stack.imgur.com/ElPI7.jpg and For my solution that is laurent expansion of 1/z^2 at z=i so I think this is wrong question but im doing partial fraction for this question and i think it can be solved thx – 서민규 Dec 07 '18 at 05:38
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given answer is not belong to this question I think – Nosrati Dec 07 '18 at 07:13
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A typo in my formula for $w$. It should read $w=(1-i)/\sqrt2$ (or some other equivalent form). – Jyrki Lahtonen Dec 07 '18 at 12:51