# How do you convert 1 - (sqrt3)i to polar form?

In polar form expressed as $2 \left(\cos 300 + i \sin 300\right)$
Let $Z = 1 - \left(\sqrt{3}\right) i$ ; Modulus |Z|=(sqrt(1^2+ (-sqrt3)^2)) =2 ; tan theta = sqrt (-3)/1 or tan theta = sqrt (-3) Argument $\theta = {\tan}^{-} 1 \left(\sqrt{- 3}\right) = - {60}^{0} \mathmr{and} {300}^{0}$.
In polar form expressed as $| Z | \cdot \left(\cos \theta + i \sin \theta\right) \mathmr{and} 2 \left(\cos 300 + i \sin 300\right)$[Ans]