How to find the intersection of complex roots, Im(z) and Re(z)?

[Cottontails]

Homework Statement

(a) Find the 6th complex roots of √3 + i.
(b) Let A={z|z^6 =√3+i} and B={z|Im(z)>0} and C={z|Re(z)>0}. Find A ∩ B ∩ C.

Homework Equations

z^6=2(cos(π/6)+isin(π/6))
r^6=2, r=2^1/6
6θ=π/6+2kπ, θ=π/36+kπ/3

The Attempt at a Solution

I've done part (a):
When k=0, z = 2^1/6(cos(π/36)+isin(π/36)),
When k=1, z = 2^1/6(cos(13π/36)+isin(13π/36)),
When k=2, z = 2^1/6(cos(25π/36)+isin(25π/36)),
When k=3, z = 2^1/6(cos(37π/36)+isin(37π/36)),
When k=4, z = 2^1/6(cos(49π/36)+isin(49π/36)),
When k=5, z = 2^1/6(cos(61π/36)+isin(61π/36)).
As for part (b) though, I am unsure for what the answer is. I understand that A={z|z^6 =√3+i} relates to part (a) however, I don't understand how you would find the intersection of A, B and C.
Please help? Thanks.

 

[grindfreak]

Ha, they're throwing a little set theory your way. Set A is every answer in a), however set B is the set of all elements that have a positive imaginary part (the ones with sin>0) and C is the set of all elements that have a positive real part (cos>0) now A ∩ B ∩ C is the intersection of all these sets and is thus the set of those in part a) (basically every element in your universe), has sin>0 AND has cos>0. Find out which ones belong to it and you've got your answer.

 

[Cottontails]

So, is it simply then just doing sinπ/36=0.0871557425... cosπ/36=0.9961946981... to find the answers to see whether they belong to the set or not?
I did that, and got that When k=0, z = 2^1/6(cos(π/36)+isin(π/36)),
When k=1, z = 2^1/6(cos(13π/36)+isin(13π/36) belong to the set of all three.
So, is the intersection then z = 2^1/6(cos(π/36)+isin(π/36)) and z = 2^1/6(cos(13π/36)+isin(13π/36)?
Moreover, could you then write the answer as A ∩ B ∩ C = z=2^1/6(cos(π/36)+isin(π/36)), z=2^1/6(cos(13π/36)+isin(13π/36)?

 

[grindfreak]

Yes that should be the answer, since both of those equations meet the criterion of the set.

 

[Cottontails]

Okay, great. Thank you very much for you help!