Question

Find [4(cos15° + isin15°)]^3 and express it in a + bi form.

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using De Moivre's theorem

Given

[ 4(cos15° + isin15° ) ]³, then

= 4³ [ cos(3 × 15°) + isin(3 × 15°) ]

= 64 (cos45° + isin45° )

= 64 ($\frac{\sqrt{2} }{2}$ + $\frac{\sqrt{2} }{2}$ i )

= 64 ($\frac{\sqrt{2} }{2}$ (1 + i) )

= 32$\sqrt{2}$ (1 + i)

= 32$\sqrt{2}$ + 32$\sqrt{2}$ i