Notam z²=t
t²+2it-4=0
Δ=(2i)²+16=-4+16=12
t1=√3-i
t2=-√3-i
z²=√3-i
Scriem √3-i si -√3-i sub forma trigonometrica.
√3-i =r(cosФ+isinФ)=2(cos11π/6+i*sin11π/6)
r=√(√3)²+1²=2
Ф=arctg(-1/√3)+2π=2π-π/6=11π/6
Calculam radacinile de ordin 2 ale numarului √3-i .
[tex]z_k= \sqrt{2} (cos \frac{ \frac{11\pi}{6} +2k\pi}{2} +isin \frac{ \frac{11\pi}{6} +2k\pi}{2})\\ z_0= \sqrt{2} (cos \frac{11\pi}{12} +isin \frac{11\pi}{12})\\ z_1= \sqrt{2} (cos \frac{23\pi}{12} +isin \frac{23\pi}{12}) [/tex]
Analog, calculam radacinile de ordin 2 ale numarului -√3-i.
