convert (-sqrt2, -sqrt6) to polar coordinates with

Question

r > 0 and 0 ≤ θ < 2π.

Answer 1

tanθ = -√6/-√2 = √3 , where θ is in quad III
θ = 180°+60° = 240° or 4π/3 radians
r = √(-√2)^2 + (-√6)^2) = √8

(-sqrt2, -sqrt6) <-----> (√8, 4π/3)