# How do I calculate residue and holomorphic range?

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?

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.
• 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? – Ana Matijanovic Apr 30 '17 at 11:41
• for a pole of order $n$ at $a$ you multiply by $(z-a)^n$. See here: en.wikipedia.org/wiki/Residue_(complex_analysis) – qbert Apr 30 '17 at 15:51
• 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 $-i$. – Ana Matijanovic Apr 30 '17 at 16:23
• 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 $-i$ – qbert Apr 30 '17 at 17:03