Correct Answer - Option 2 :
\(2\left( {\cos \left( {\frac{{2\pi }}{3}} \right) + isin\left( {\frac{{2\pi }}{3}} \right)} \right)\)
CONCEPT:
Point P is uniquely determined by the ordered pair of real numbers (r, θ), called the polar coordinates of the point P.
P represent the nonzero complex number z = x + iy.
Here \(r = \sqrt {{x^2} + {y^2}} = \left| z \right|\) is called the modulus of the given complex number.
The argument of Z is measured from positive x-axis only.
Let z = r (cosθ + i sinθ) is polar form of any complex number then following ways are used while writing θ for different quadrants –
For the first quadrant, \({\rm{\theta }} = {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)
For the second quadrant \({\rm{\theta }} = {\rm{\pi }} - {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)
For the third quadrant \({\rm{\theta }} = - {\rm{\pi }} + {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)
For the fourth quadrant \({\rm{\theta }} = - {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)
CALCULATION:
Given complex number \(z = \frac{{ - 4}}{{1 + i\sqrt 3 }}\)
On rationalization we get –
\(z = \frac{{ - 4}}{{1 + i\sqrt 3 }} \times \frac{{1 - i\sqrt 3 }}{{1 - i\sqrt 3 }} = \frac{{ - 4\left( {1 - i\sqrt 3 } \right)}}{{{1^2} - {{\left( {i\sqrt 3 } \right)}^2}}} = \frac{{ - 4\left( {1 - i\sqrt 3 } \right)}}{4} = - 1 + i\sqrt 3 \)
rcosθ = -1, rsinθ = √3
By squaring and adding, we get –
\({{\rm{r}}^2}\left( {{\rm{co}}{{\rm{s}}^2}{\rm{\theta }} + {\rm{si}}{{\rm{n}}^2}{\rm{\theta }}} \right) = 1 + 3\)
∴ r = 2
\(\Rightarrow cos\theta = \frac{{ - 1}}{r} = \frac{{ - 1}}{2}\) and \(sin\theta = \frac{{\sqrt 3 }}{r} = \frac{{\sqrt 3 }}{2}\)
Since it is in the second quadrant, \(\theta = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3}\)
So, on comparing with z = r (cosθ + i sinθ), we can write as
\(2\left( {\cos \left( {\frac{{2\pi }}{3}} \right) + i\;sin\left( {\frac{{2\pi }}{3}} \right)} \right)\) 
