Evaluation of the integral $\oint _{ C }^{ }{ \frac { 2{ z }^{ 2 }+5 }{ { \left( z+2 \right) }^{ 3 }\left( { z }^{ 2 }+4 \right) { z }^{ 2 } } dz } $

Question

It seems that to find residue and then applying the Residue Theorem" is too long . I'd like to know is there a briefer way to evaluate this integral apart from applying the Residue Theorem? $$\oint _{ C }^{ }{ \frac { 2{ z }^{ 2 }+5 }{ { \left( z+2 \right) }^{ 3 }\left( { z }^{ 2 }+4 \right) { z }^{ 2 } } dz } $$
where contour here is $\left| z-2i \right| =6$

Answer 1

Yes, there is an easier way here. By the residue theorem, your contour integral equals $2\pi i$ times the sum of the residues inside $C.$ But there are no singurities beyond those inside $C.$ Thus any contour containing $C$ can make the same claim. So if $R$ large, say $R>100,$ we have

$$\int_C { \frac { 2{ z }^{ 2 }+5 }{ { \left( z+2 \right) }^{ 3 }\left( { z }^{ 2 }+4 \right) { z }^{ 2 } } dz } = \int_{|z|=R}{ \frac { 2{ z }^{ 2 }+5 }{ { \left( z+2 \right) }^{ 3 }\left( { z }^{ 2 }+4 \right) { z }^{ 2 } } dz }.$$

Now consider the contour integral on the right. In absolute value, it is no more than $2\pi R$ times the maximum of the integrand in absolute value. This maximum, as a function of $R,$ is on the order of $\dfrac{ R^2}{R^3\cdot R^2\cdot R^2}.$ It follows that the limit of the integral on the right, as $R\to \infty,$ is $0.$ Well, if the limit of something that is constant is equal to $0,$ then that something is constantly $0.$ Hence the contour integral on the left equals $0.$