Question

z=(1+i)^{2017}+(1-i)^{2017} .Claculati |x|

Answer 1

Presupun ca  lzl se  cere
Fie  z1=1+i. Il  scrii  sub  forma  trigonometrica
lz1l=√(1²+1²)=√2
cosα=1/√2=√2/2  sinα=√2/2=>  α=π/4
z1=√2*(cosπ/4+isinπ/4)
aplici  formula  lui Moivre  si  determini
z^2017==(√2)^2017*(cos2017π/4+isin2017π/4)
Z2=1-i  conjugatul  lui  z1
z2=(√2/2)*(cosπ/4-i sinπ/4)
z2^2017=(√2/2)^2017*(cos2017π4-isin2017π/4)
Z=√2^2017*(cos2017π/4+isin2017π/4)+√2/2^2017*(cos2017π/4-isin2017π/4)
z=2*√2^2017 cos2017π/4
lzl=2*√2^2017*lcos2017π/4l=2*√2^2017*lcos(2016π/4+π/4)=2*√2^2017*cosπ/4
2*√2^2017*√2/2=√2^1009