Convert (-2/sqrt[3] , 2) to polar coordinates

If you have these conditions:

1. (r ,theta) where r > 0 
2. (r, theta) where r < 0

The solution would be:

r = sqrt(x^2 + y^2) 

t = arctan(y/x) 

r = sqrt(12 + 4) = sqrt(16) = +/- 4 

t = arctan(2 / -2sqrt(3)) = arctan(-1 / sqrt(3)) = 5pi/6 , 11pi/6 

1) 
r > 0 
(4 , 11pi/6) 

2) 
r < 0 
(-4 , 5pi/6)


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