[tex] 4) \left(\dfrac{z+2i}{z-i}\right)^3- \left(\dfrac{z+2i}{z-i}\right)^2+ \left(\dfrac{z+2i}{z-i}\right)-1=0\\
\left(\dfrac{z+2i}{z-i}\right)^2 \left(\left(\dfrac{z+2i}{z-i}\right)-1\right)+\left(\left(\dfrac{z+2i}{z-i}\right)-1\right)=0\\
\left(\left(\dfrac{z+2i}{z-i}\right)-1\right)\left(\left(\dfrac{z+2i}{z-i}\right)^2+1\right)=0\\
\text{Le egalam pe ambele cu 0:}\\
1)\dfrac{z+2i}{z-i}=1\\
z+2i=z-i\\
2i=-i(F)\\
2)\left(\dfrac{z+2i}{z-i}\right)^2=-1\\
\text{Si de aici se deduc doua cazuri:}
[/tex]
[tex]a)\left(\dfrac{z+2i}{z-i}\right)=-i\\
z+2i=-iz-1\\
z(1+i)=-2i-1\\
z=\dfrac{-1-2i}{1+i}\\
\\
b)\left(\dfrac{z+2i}{z-i}\right)=i\\
z+2i=iz+1\\
z(1-i)=1-2i\\
z=\dfrac{1-2i}{1-i}[/tex]
[tex]1) \text{Se foloseste forma trigonometrica:}\\
z^{10}=-1\\
\text{Avem ca:} \ r=1\\
\sin \alpha=0\\
\cos \alpha =-1, deci\ \alpha=\pi\\
z^{10}=\cos \pi+ i\sin \pi\\
\text{Folosind formulele lui Moivre obtinem:}\\
z_{k}=\cos \dfrac{\pi +2k\pi}{10}+i\sin \dfrac{\pi +2k\pi}{10};\ k=\overline{0,9}[/tex]
