Question

z=(1+i)^{2017}+(1-i)^{2017}.Ultima cifra a lui |z|

Answer 1

z=(√2(cos π/4+isinπ/4))^2017+ (√2(cos 7π/4+isin7π/4))^2017

studiem argumentele celor doua numer complexe insumate
2017π/4=2016π/4+π/4=504π+π/4= (2π)*252+π/4=π/4

2017*7π/4=14119π/4=14116π/4+3π/4=3529π+3π/4=
3528π+π+3π/4=1764*2π+π+3π/4=7π/4

atunci
(√2(cos π/4+isinπ/4))^2017+ (√2(cos 7π/4+isin7π/4))^2017=

√2^2017(cosπ/4+isinπ/4+cos7π/4+isin7π/4)= √2^2017*2cos7π/4=
 √2 ^ 2017*2*1/√2= 2 * √2^2016=2*2^1008=2^1009
Z este un nr real pozitiv si atunci |z|=z=2^1009

deci este un numar real, o putere a lui 2

U(2^n) se repeta din 4 in 4
 2009 dpdv al impartirii la 4 , perioada functiei U (2^n), estede forma 4k+1
deci U( 2^1009) =U (2^1) =U(2)=2