Concept:

The polar form of the complex number z = (x + iy) is 

⇒ z = r (\(\rm \cos\;\theta + i sin\;\theta\)) , where r = \(\sqrt {x^2+y^2}\) and \(\rm \theta = tan^{-1}(\dfrac y x)\)

De - Moivers thereom:

Let z = r (\(\rm \cos\;\theta + i sin\;\theta\)) be the polar form of the complex number z = x + iy.

then the value of zⁿ = [r (\(\rm \cos\;\theta + i sin\;\theta\))]ⁿ is rⁿ(\(\rm \cos\;n\theta + i sin\;n\theta\))

Calculations:

Given, complex number is  \((-1+i\sqrt{3})^{48} \)

Consider z = (x + iy) =\((-1+i\sqrt{3}).\) 

⇒ x = 1 and y = \(\sqrt 3\)

The polar form of the complex number is 

⇒ z = r (\(\rm \cos\;\theta + i sin\;\theta\)) , where r = \(\sqrt {x^2+y^2}\) and \(\rm \theta = tan^{-1}(\dfrac y x)\)

Here r = \(\sqrt {1^2+(\sqrt3)^2}\) = 2 and \(\rm \theta = tan^{-1}(\dfrac {\sqrt3}{1})\) = \(\dfrac{\pi}{3}\)

⇒ z = 2 (\(\rm \cos\;\dfrac {\pi}{3} + i sin\;\dfrac{\pi}{3}\))

⇒ z⁴⁸ = [2 (\(\rm \cos\;\dfrac {\pi}{3} + i sin\;\dfrac{\pi}{3}\))]⁴⁸

⇒ z48 = 2⁴⁸ (\(\rm \cos\;(48\times\dfrac {\pi}{3}) + i sin\;(48\times\dfrac{\pi}{3})\)

⇒ z48 = 248 [\(\rm \cos\;(16 {\pi}) + i sin\;(16{\pi})\)]

⇒ z48 = 248 [\(\rm [1 + i(0)]\)]

⇒ z48 = 248

Hence, the value of \((-1+i\sqrt{3})^{48} \) = 248