How do I determine the poles of 1/(z^6 + 1) above the y-axis?

[Lavabug]

Kind of stuck (embarrassingly) on determining what poles of the function:

1/(z^6 + 1)

lie above the y-axis (I'm solving a contour integral using the residues theorem).

What's the easiest way to do this? Normally I'd write z as e^(theta + 2kpi)/6 , where theta is the angle that -1 forms with the first quadrant (pi) and solve for 6 values of k but I'm pretty confused on this problem.

Isn't there a more direct way to get all 6 roots of z^6 + 1 = 0?

 

[tiny-tim]

Hi Lavabug!

(have a theta: θ and a pi: π and try using the X² icon just above the Reply box )

Normally I'd write z as e^(theta + 2kpi)/6 , where theta is the angle that -1 forms with the first quadrant (pi) and solve for 6 values of k but I'm pretty confused on this problem.

I don't understand …

that's π/6, 3π/6, 5π/6, etc …

what's wrong with that? ​

 

[Lavabug]

Hi Lavabug!

(have a theta: θ and a pi: π and try using the X² icon just above the Reply box )


I don't understand …

that's π/6, 3π/6, 5π/6, etc …

what's wrong with that? ​

Thanks. I'll make an effort to use learn the Tex in following posts, as you can imagine at this time of year I'm in a hurry! :p

Yeah those are the exponents I get that have projections above the y axis, but they're really ugly numbers (solved for them using Euler's formula) to plug into the residues formula. Something tells me I'm doing something wrong, shouldn't my poles be +-i raised to the power of something?

Also how would I decompose the denominator into perfect squares so the (z -z0)^m terms cancel when I calculate the residue? I've only dealt with 2nd, 4th order polys in the denominator but this one has me scratching my head.

 

[SteamKing]

You are trying to find solutions to z^6 + 1 = 0. This is similar to solving z^6 - 1 = 0, except the solutions to your equation will be rotated. The roots of z^6 + 1 will plot as the vertices of a hexagon, and one vertex will be positioned at (0, i) and one at (0, -i). You should review the topic "Roots of unity" for more details.