Contour Integral: how to get the order of those poles?

[degs2k4]

Homework Statement

The following function :

a) $$f(z) = \frac{1}{z^6 + 1}$$

has simple poles on :

$$z_1 = e^{pi/6 i}, z_2 = e^{3pi/6 i}, z_3 = e^{5pi/6 i}$$

I know how to get the poles, but how could I demonstrate they are simple (order 1) ?
I tried to write the Laurent series centered in the pole, but I got nowhere...
Is there any general way to determine the order of a pole without using Laurent ?

Thanks.

 

[lanedance]

expanding the denominator could help, not also that theta = 7 pi/6, 9 pi/6 and 11 pi/6 are also poles

 

[gomunkul51]

Your function always could be written as 1 / [(z-a)(z-b)(z-c)(z-d)(z-e)(z-f)]
where a,b,c,d,e,f are the roots of the polynomial z^6 - 1 = 0.

The poles are simple poles, because there are no repeated roots.
you have 6 complex roots to the polynomial z^6 - 1 = 0.
I assume you want to do an integral using the "residue theorem"
so you correctly chose 3 of then that are in the positive imaginary direction.

*if your function was 1 / [(z-2)(z-3)^3] then you've had one simple pole at z=2 and one pole of order 3 at z=3.

this is a general explanation.

 

[degs2k4]

not also that theta = 7 pi/6, 9 pi/6 and 11 pi/6 are also poles

Your function always could be written as 1 / [(z-a)(z-b)(z-c)(z-d)(z-e)(z-f)]
where a,b,c,d,e,f are the roots of the polynomial z^6 - 1 = 0.

The poles are simple poles, because there are no repeated roots.
.

Thanks! I think I got it now.