Evaluate

Algebra Level 3

Evaluate [6(cos 30° + i sin 30°)]^{2}

36 + 363√i 18 + 183√i 36 - 183√i 18 - 363√i

2 solutions

J. Fabio Ramírez Pérez
Jun 9, 2015

Using DeMoivre's Theorem

[r(cos θ+j sin θ)]^n =r^n (cos nθ+j sin nθ)

we should get

36(cos60°+i sin60°)
18+i 18 sqrt(3)

Aditya Raj
Feb 4, 2015

Solution:

Let Z = 6(cos 30° + i sin 30°)

Z 2 = 6 6[cos (2 · 30°) + i sin (2 · 30°)] [Use De MoivreÃ¢â‚¬â„¢s theorem.]

= 36[cos 60° + i sin 60°]

= 36[12 + 32 i] [Use cos 60° = 1 / 2, sin 60° = 32.]

= 18 + 183i.

Shouldn't the options have 18 root3 i

Abhishek Garg - 6 years, 4 months ago