Question

sa.se rezolve în mulțimea numerelor complexe:​

Answer 1

Răspuns:

Explicație pas cu pas:

z⁴=(2-i)/[2+√3+i(2√3-1)]=[2+√3-i(2√3-1)]·(2-i)/[(2+√3)²+(2√3-1)²]

[2+√3-i·(2√3-1)]·(2-i)=4+2√3-i·(4√3-2)-i·(2+√3)+i²·(2√3-1)=

=4+2√3-i·(4√3-2+2+√3)-2√3+1=5-i·5√3

(2+√3)²+(2√3-1)²=4+4√3+3+12-4√3+1=20

⇒ z⁴=(5-i·5√3)/20=1/4-i√3/4=(1/2-i√3/2)/2

Daca z=r·(cosФ+i·sinФ), conform formulei lui Moivre avem:

zⁿ=rⁿ·(cos(n·Ф)+i·sin(n·Ф)) ⇒ z⁴=r⁴·(cos(4·Ф)+i·sin(4·Ф))

Dar |z⁴|=|z|⁴=r⁴=|(1/2-i√3/2)/2|=1/2 ⇒ $r=\sqrt[4]{2}$

cos(4·Ф)=1/2 si sin(4·Ф)=-√3/2

⇒ 4·Ф=-π/3+2·k·π ⇒ Ф=-π/12+k·π/2

⇒ z=$\sqrt[4]{2}$·(cos(-π/12+k·π/2)+i·sin(-π/12+k·π/2)) , k=0,1,2,3