Question

26-1 si 4 daca puteti
mersi

Answer 1

[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]