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Calculate residue and holomorphic range of this function. $f(z)=\frac{z-i}{(z^2+1)^3}$

My idea and problem:

Singularities: $z_0=i,-i$

$Res(f,i)=\lim_{z \to i} (z-i)\frac{(z-i)}{(z^2+1)^3} $ I used two times L'Hospital's rule on this formula, but I still cant get the result. Where is my mistake and how do I determine holomorphic range?

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Your formula is valid for poles of order 1, note that you have $$ f(z)=\frac{z-i}{(z^2+1)^3}=\frac{1}{(z+i)^3(z-i)^2} $$ and thus you have poles of order 2 and 3. I'll let you take it from here.

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  • $\begingroup$ Here is what I tried for residue near $i$: $Res(f,i)=\lim_{z \to i} \frac{d}{dz}(z-i)\frac{(1)}{(z+i)^3(z-i)^2} =\lim_{z \to i} \frac{(-5z+i)}{(z-i)^3(z+i)^4}$ And when I try to use L'Hospital's rule here,I will again have term $(z-i)$ in denominator? $\endgroup$ – Ana Matijanovic Apr 30 '17 at 11:41
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    $\begingroup$ for a pole of order $n$ at $a$ you multiply by $(z-a)^n$. See here: en.wikipedia.org/wiki/Residue_(complex_analysis) $\endgroup$ – qbert Apr 30 '17 at 15:51
  • $\begingroup$ Great, I finally did it. Can you just help me with type of singularity ofr $-i$? For $i$ I think it is pole, cause $\lim_{z \to z_0} f(z)=\infty $. But I have trouble with calculating limit for $-i$. $\endgroup$ – Ana Matijanovic Apr 30 '17 at 16:23
  • $\begingroup$ It is absolutely a pole, but what you said holds for essential singularities as well. Make sure you know the difference. The computation is nearly identical for $-i$ $\endgroup$ – qbert Apr 30 '17 at 17:03

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