Question

O valor de (1+i)^4040 é: a)i b)2i c) c?(2ˆ2020)i d)-(2ˆ2020)i e)2ˆ2020

Answer 1

$(i+i)^2=1^2+2i+i^2=1+2i-1=2i\\\\(1+i)^8=((1+i)^2)^4=(2i)^4=16i^4=16\\\\(1+i)^{4040}=((1+i)^8)^{505}=16^{505}=\boxed{2^{2020}}$