Now that you have figured out the set A(which is not entirely correct,please look at it again),choose only those elements which have their real and imaginary parts greater than 0.In short both cos and sin terms should be positive.
Let A = {z|z^{6} = '3 + i} and B = {z|Im(z) > 0} and C = {z|Re(z) > 0
(where '3 means the square root of 3)
.. I found the z^{6} to be: (using k = 0,1,2,3,4,5 in that order)
z_{1}= z^{1/6}(cos pi/6 + isin pi/6)
z_{2}= z^{1/6}(cos 2/pi + isin 2/pi)
z_{3}= z^{1/6}(cos 5pi/6 + isin 5pi/6)
z_{4}= z^{1/6}(cos 7pi/6 + ison 7pi/6)
z_{5}= z^{1/6}(cos 9pi/6 + isin 9pi/6)
z_{6}=z^{1/6}(cos 11pi/6 + isin 11pi/6)
I've been asked to find the 'intersecting' of sets A, B, and C.. But not really sure how I'm meant to compapre real, imaginary and complex numbers together..
but does that mean I pretty much write my answer out to set A again..
Literally :
z1= z1/6(cos pi/6 + isin pi/6)
z2= z1/6(cos 2/pi + isin 2/pi)
z3= z1/6(cos 5pi/6 + isin 5pi/6)
z4= z1/6(cos 7pi/6 + ison 7pi/6)
z5= z1/6(cos 9pi/6 + isin 9pi/6)
z6=z1/6(cos 11pi/6 + isin 11pi/6)
Go through this once:Complex number - Wikipedia, the free encyclopedia
Intersection of Sets B and C gives you the first quardant.Consider only those z which lie in first quadrant and satisfy condition of set A.