(x, y) = (-1, -1) 

∴ r = \(\sqrt {x^2+y^2} = \sqrt{1+1}=\sqrt2\)

tan θ = \(\frac{y}{x} = \frac {-1}{-1}=1\)

∴ tan θ = tan π/4

Since the given point lies in the 3rd quadrant, 

tan θ = tan (π + π/4) …[∵ tan (n + x) = tan x] 

∴ tan θ = tan (5π/4)

∴ θ = (5π/4) = 225° 

∴ the required polar co-ordinates are ( \(\sqrt2\), 225°).

(x, y) = (- √ 3, 1) 

∴ r = \(\sqrt{x^2+y^2} = \sqrt{3+1}=\sqrt4=2\)

tan θ = \(\frac{y}{x} = \frac {1}{-\sqrt3}=-tan \frac{\pi}{6}\)

Since the given point lies in the 2nd quadrant, 

tan θ = tan (π-π/6) …[∵ tan (π – x) = – tan x] 

∴ tan θ = tan (5π/6)

∴ θ = (5π/6)= 150° 

∴ the required polar co-ordinates are (2, 150°)

(x, y) = ( 1,√3)

∴ r =\(\sqrt{x^2+y^2} = \sqrt{1+3} = \sqrt4 = 2\)

tan θ = \(\frac{y}{x} = \frac {\sqrt3}{1} =\sqrt3\)

Since the given point lies in the 1st quadrant, 

θ = 60° …[∵ tan 60° = √3] 

∴ the required polar co-ordinates are (2, 60°).

(x, y) = (5, 5) 

∴ r = \(\sqrt{x^2+y^2} = \sqrt{25+25}\)

= \(\sqrt{50} = 5\sqrt{2}\)

tan θ = y/x = 5/5 = 1 

Since the given point lies in the 1st quadrant, 

θ = 45° …[∵ tan 45° = 1] 

∴ the required polar co-ordinates are ( \(5\sqrt{2}\), 45°).