Laurent Expansion Problem (finding singularities)

[Appledave]

Homework Statement

Find all Laurent expansion of the function f(z) = 1/(z(8(z^3)-1)) with centre z = 0.

The Attempt at a Solution

I tried to find all the singularities and came up with z = 0, z = 1/2, z = (1/2)exp((n*pi*i)/3)
where n = +-2,+-4,+-6... . But according to the solution n can only be 2 and 4, what am I missing?

 

[Count Iblis]

n = 4 and n = -2 are the same (and n = -4 is the same as n = 2) n = 6 is the same as n = 0 which yields z = 1/2.

 

[Appledave]

exp((n*pi*i) = cos(n*pi) + i*sin(n*pi) = 1 + i*0 = 1 for n = +-2,+-4,+-6,+-8... isn't it? Wouldn't this make z = (1/2)exp((n*pi*i)/3) yield z = 1/2 for all n = +-2,+-4,+-6,+-8... ?
(making all n = +-2,+-4,+-6,+-8... equal, hence my confusion over why only n=2 and n=4 were included in the solution).

 

[Count Iblis]

There is a factor 1/3 in the exponential.

 

[Appledave]

But that factor disappears because you take z to the power of three in the function
(f(z) = \frac{1}{z(8z^{3}-1)})

Edit: nvm this post, I got it. thanks :)