Question

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

Answer 1

`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))))`

Answer 2

`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`