Question

[(1+i)/(1-i)]^102, i= raiz de -1 tem módulo igual a?

Answer 1

[(1+i)/(1-i)]^102 = 
[(1+i).(1+i) / (1-i).(1+i)]^102 = 
[(1²+2i+i²) / 1² - i²]^102 = 
[(1+2i-1) / (1 -(-1)]^102 = 
[(2i)/(1+1)]^102 = 
[2i/2]^102 = 
i^102...
102:4 = 25 e sobram 2
o resto (2) será o expoente de i, logo...
i^102 = i² = -1