
Now, separating real and complex part , we get
\(\frac{1-\sqrt3}{2}\) = rcosθ……….eq.1
\(\frac{1-\sqrt3}{2}\) = rsinθ…………eq.2
Squaring and adding eq.1 and eq.2, we get
2 = r2
Since r is always a positive no., therefore,
r = √2,
Hence its modulus is √2.
Now, dividing eq.2 by eq.1 , we get,

Therefore the θ lies in second quadrant.
As Tanθ = \(\frac{1+\sqrt3}{1-\sqrt3}\), therefore θ = 7π/12
Representing the complex no. in its polar form will be
Z = √2{cos(7π /12)+i sin(7π /12)}