Question

Determine o número natural n tal que (2i)^n + (1+i)^2n +16i=0

Answer 1

(2i)^n + (1+i)^2n +16i=0
(2i)^n + ((1+i)^2)^n) + 16i = 0
(2i)^n + (1 +2i + i²)^n + 16i = 0               i² = -1 
(2i)^n + (1 +2i  -1)^n + 16i = 0
(2i)^n + (2i)^n + 16i = 0
(2i)^n + (2i)^n = -16i                      

n vale 3 , 
(2i)³ + (2i)³ = -16i
8i³ + 8i³ = - 16i
8.-i + 8.-i = - 16i
-8i - 8i = -16i
-16i = -16i , bons estudos