How to solve the integral using the residue theorem int_(|z|=2)z^10/(z^11+1)dz ?

2 Answers
May 9, 2017

int_(abs z =2) z^10/(z^11+1) dz = 2pi i sum_(k=0)^10 e^((9ipi)/11(2k+1))/(prod_(j!=k) (1-e^((2ipi)/11(j-k))))

Explanation:

Consider the path of integration as the simple closed circle of radius abs z = 2 and center in the origin.

Then based on the residue theorem:

int_(abs z =2) z^10/(z^11+1) dz = 2pi i sum_k Res_f" (z_k)

where z_k are the singularities of f(z) = z^10/(z^11+1) falling inside the circle.

Now f(z) is a rational function, therefore it is holomorphic everywhere except where the denominator vanishes, which means that its singularities are given by the equation:

z^11+1= 0

so the z_k are the roots of 11-th order of -1:

z_k = e^((ipi)/11(2k+1)) " for " k = 0,1,...,10

and they are all poles of order 1.

Then:

Res(z_k) = lim_(z->z_k) (z-z_k)f(z)

Now if we express the function as:

f(z) = z^10/(z^11+1) = z^10/(prod_(j=0)^10 (z-z_j))

then we have:

(z-z_k)f(z) = (z-z_k) z^10/(prod_(j=0)^10 (z-z_j)) = z^10/(prod_(j!=k) (z-z_j)

which is continuous for z=z_k, so:

Res(z_k) = z_k^10/(prod_(j!=k) (z_k-z_j)) = e^((10ipi)/11(2k+1))/(prod_(j!=k) (e^((ipi)/11(2k+1))-e^((ipi)/11(2j+1))))

Res(z_k) = e^((10ipi)/11(2k+1))/(e^((ipi)/11(2k+1))(prod_(j!=k) 1-e^((2ipi)/11(j-k)))) = e^((9ipi)/11(2k+1))/(prod_(j!=k) (1-e^((2ipi)/11(j-k))))

and finally:

int_(abs z =2) z^10/(z^11+1) dz = 2pi i sum_(k=0)^10 e^((9ipi)/11(2k+1))/(prod_(j!=k) (1-e^((2ipi)/11(j-k))))

May 9, 2017

2pi i

Explanation:

As |z|=2 represents a circle of radius 2 centred on the origin, then we can write the integral as:

I = oint_C \ z^10/(z^11+1) \ dz " " where C is the circle |z|<2

The integrand has singularities when z^11+1=0 which corresponds to the 11 simple poles given by z=root(11)(-1). All of these poles lie within the circle C and there are no other poles.

Thus, by the residue theorem:

I = 2pii \ xx {"sum of residues within "C}

So in order to evaluate the integral we just need to find the residues at the 11 poles, Using Cauchy's integral formula we find all poles have a residue of 1/11

Thus

I = 2pi i \ * 1/11 * 11 = 2pi i