Find $\int_\Gamma\frac{2z+j}{z^3(z^2+1)}\mathrm{d}z$ where $Γ:|z-1-i| = 2$ pls, some ideas for integral solution (residue theory)?
$$\int_\Gamma\dfrac{2z+j}{z^3(z^2+1)}\mathrm{d}z$$
Where $Γ:|z-1-i| = 2$ is positively oriented circle.
Thx, for help!
 A: For all $z\in \mathbb C\setminus \{-i, 0,i\}$ set $f(z)=\dfrac{2z+i}{z^3(z^2+1)}$.
It holds that
$$\dfrac{2z+i}{z^3(z^2+1)}=\dfrac i {z^3}+\dfrac 2{z^2}+\dfrac {-i}z+\dfrac{3i}{2(z-i)}+\dfrac {-1}{2(z+i)}.$$
(Courtesy of WA).
So you can find the laurent series around $0$ of $f$. You don't have to compute it to tell that the coefficient of $z^{-1}$ is $-i$ (why?). This coefficient is also $\text{Res}(f,0)$.
The residues at $i$ and $-i$ are first order poles and so they are easy to find, but you only need one of them, because one lies outside the circle.
Making the assumption that $\Gamma$ goes around the circle only once, the residue theorem yields
$$\int _\Gamma f=2\pi i\left(\text{Res}(f,0)+\text{Res}(f,i)\right).$$
A: So, my solution is:
Is it correct ??
pole z1 = 0 ( order 3 pole );
pole z2 = -i ( simple pole );
pole z3 = i ( simple pole );
z1 : |0-1-i| = sqrt(2) < 2 => in circle;
z2 : |-i-1-i| = sqrt(5) > 2 => not in circle;
z3 : |i-1-i| = 1 < 2 => in circle;
$$
\underset{z=0}{res}\frac{2z+i}{z^3(z^2+1)}=-i
$$
$$
\underset{z=i}{res}\frac{2z+i}{z^3(z^2+1)}=\frac{3i}{2}
$$
$$
\int_\Gamma\dfrac{2z+i}{z^3(z^2+1)}\mathrm{d}z = 2\pi i(\underset{z=0}{res} f(z) + \underset{z=i}{res} f(z)) = 2\pi i(-i+\dfrac{3i}{2}) = -\pi 
$$
