Simplify by using De-Moivre’s theorem.

Question 

Simplify `\left( \sqrt 3 + 3i \right)^{31}`  by using De-Moivre’s theorem.             

Solution:

De-Moivre’s theorem

`z = r^n\left(\cos n\theta  + i\sin n\theta  \right)`  ------- (1)

Let

`z = \left( \sqrt 3 + 3i \right)^{31}`

`n = 31`

`\| z| = r = \sqrt {\left( \sqrt 3  \right)^2 + \left( 3 \right)^2}`

`\ = \sqrt {3 + 9}`

`\ = \sqrt {12}`

`\arg z = \theta = \tan ^{ - 1}\left(\frac{y}{x} \right)`

`\= \tan ^{ - 1}\left(\frac{\sqrt 3 }{3} \right)`

`\= \tan ^{ - 1}\left(\frac{\sqrt 3}{\sqrt 3 \sqrt 3 } \right)`

`\= \tan ^{ - 1}\left( \frac{1}{\sqrt 3 } \right)`

`\= \frac{\pi }{6}`

Now

`\left( \sqrt 3 + 3i \right)^{31} = \sqrt {12} ^{31}\left( \cos 31\left( \frac{\pi }{6} \right) + i\sin 31\left( \frac{\pi }{6} \right) \right)`

`\left( \sqrt 3  + 3i \right)^{31}= \sqrt {12} ^{31}\left( \cos \frac{31\pi }{6} + i\sin \frac{31\pi }{6} \right)`

`\left( \sqrt 3 + 3i \right)^{31} = \sqrt {12} ^{31}\left( cis\frac{31\pi }{6} \right)`