(1-i)z^6=1-i√3
1-i√3=2(1/2-i√3/2)=2(cos5π/3+isin5π/3)
pt ca |1-i√3|=√(1²+(-√3)²)=√(1+3)=√4=2 si modulul se da factor comun fortat
1-i=√2(1/√2-i√/2)=√2(cos7π4+isin7π/4) pt ca;
|1-i|=√(1²+(-1)²)=√2 si modulul se da factor comun fortat
pentru unghiuri, vezi figura atasata
atunci
z^6=(1-i√3):(1-i)=
[2(cos5π/3+isin5π/3)]:[√2(cos7π/4+isin7π/4)]=aplic formula de impartire a nr.complexe scrise trigonometric=
z^6=√2[cos(5π/3-7π/4)+isin(5π/3-7π/4)]=?
cum 5π/3-7π/4=20π/12-21π/12=-π/12
z^6=√2[cos(-π/12)+isin(-π/12)]=√2[cos(23π/24) +isin(23π/24)], numar complex scris sub forma trig, avand modul √2 si argument 23π/24∈[0;2π]
am folosit -π/12=2π-π/12=24π/12-π/12=23π/12
deci
z^6=√2[cos(23π/24) +isin(23π/24)]
avem formula radacinilor de ordin n ale unui un nr.complex scris sub forma trigonometrica
z^n=modul (cosα+isinα)
solutiile sunt
zk= radical indice n din modul* [cos((α+2kπ)/n)+isin(α+2kπ)/n)]
unde n este indicele=puterea , iar k=0,1,2,...,n-1
in cazul nostru
n=6
|z^6|=√2
atunci
|zk|=
radical indice 6din(√2)= radical indice 12 din2 = 2^(1/12)
iar, cu α=23π/24, argumentele sunt:
(23π/24+2kπ)/6=23π/144+2kπ/6=(23π+48kπ)/144
cu k=0,1,2...,5
atunci cele 6 radacini de ordinul 6 sunt :
z1=2^(1/12)*[cos23π/144+isin23π/144]
z2=2^(1/12)*[cos71π/144+isin71π/144
z3=2^(1/12)*[cos119π/144+isin119π/144]
z4=2^(1/12)*[cos167π/144+isin167π/144]
z5=2^(1/12)*[cos215π/144+isin215π/144]
z6=2^(1/12)*[cos263π/144+isin263π/144]
se observa ca 263π<144*2π=288π deci x cele 6 radacini au argumentele cuprinse in intervalul [0;2π]
