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The polar form of -√3 + i will be –
1. \(\rm 2 e^\frac{\pi }{6}\)
2. \(\rm e^\frac{5\pi }{6}\)
3. \(\rm 2e^\frac{5\pi }{6}\)
4. \(\rm 2e^\frac{7\pi }{6}\)

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Correct Answer - Option 3 : \(\rm 2e^\frac{5\pi }{6}\)

CONCEPT:

If 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 the positive x-axis only.

Let z = r (cos θ + i sin θ) is a 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 \({\rm{\theta }} = - {\rm{\;ta}}{{\rm{n}}^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)

Note: The polar form z = r (cosθ + i sinθ) is abbreviated as r.cisθ.

CALCULATION:

Given \(Z = - \sqrt 3 + i\)

\(x = - \sqrt 3 ,\;y = 1\)

\(r = \sqrt {{{\left( { - \sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2}} = 2\)

\(tan\theta = \frac{1}{{ - \sqrt 3 }} \Rightarrow \theta = - {\tan ^{ - 1}}\frac{\pi }{6}\)

Here the reference angle and for θ is 30 °. Since the complex number is in the second quadrant –

\(\therefore \theta = \pi - \frac{\pi }{6} = \frac{{5\pi }}{6}\)

\(\therefore Z = - \sqrt 3 + i \)

\(\rm 2e^\frac{5\pi }{6}\)

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