Use De Moivre’s Theorem to compute the following: [3(cos27+isin27)}^5 Please explain step by step how it works!

Question

Use De Moivre’s Theorem to compute the following:
`[3(cos27+isin27)}^5` Please explain step by step how it works!

Answer 1

hello:
Use De Moivre’s Theorem :
(3(cos27 +isin27)^5 = 3^5( cos(27 × 5) +isin(27 × 5))
= 3^5 ( cos(135)+i sin(135))
= 3^5(-√2/2+i √2/2)
because : cos(135) = -√2/2 and sin(135) = √2/2
(3(cos27 +isin27)^5 = (- 3^5√2/2)+ i ( 3^5√2/2) ...(form : a+ib when
a= (- 3^5√2/2) and b = ( 3^5√2/2)