Find $\int_\Gamma\frac{2z+j}{z^3(z^2+1)}\mathrm{d}z$ where $Γ:|z-1-i| = 2$

Question

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!

Answer 1

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;

$$ \underset{z=0}{res}\frac{2z+i}{z^3(z^2+1)}=-i $$ $$ \underset{z=i}{res}\frac{2z+i}{z^3(z^2+1)}=\frac{3i}{2} $$ $$ \int_\Gamma\dfrac{2z+i}{z^3(z^2+1)}\mathrm{d}z = 2\pi i(\underset{z=0}{res} f(z) + \underset{z=i}{res} f(z)) = 2\pi i(-i+\dfrac{3i}{2}) = -\pi $$

Answer 2

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

$$\int _\Gamma f=2\pi i\left(\text{Res}(f,0)+\text{Res}(f,i)\right).$$