[tex]\it a)\ \ z^2+4z+4-2i=0 \Rightarrow (z+2)^2-2i=0\ \ \ \ \ \ (1)\\ \\ Fie\ 2i=(a+bi)^2 \Rightarrow 2i=a^2+2abi-b^2 \Rightarrow \begin{cases}\ \it 2abi=2i \Rightarrow ab=1 \Rightarrow a=\dfrac{1}{b}\ \ (2)\\ \\ a^2-b^2=0 \Rightarrow a^2=b^2\ \ \ (3)\end{cases}\\ \\ \\ (2),\ (3) \Rightarrow \dfrac{1}{b^2}=b^2 \Rightarrow b^4=1 \Rightarrow b=\pm1\stackrel{(2)}{\Longrightarrow}\ a=\pm1[/tex]
[tex]\it Prin\ urmare, \ 2 i=(-1-i)^2=(1+i)^2, \ iar\ ecua\c{\it t}ia\ \ (1)\ devine:\\ \\ (z+2)^2-(1+i)^2=0 \Rightarrow (z+2-1-i)(z+2+1+1)=0 \Rightarrow \\ \\ \Rightarrow (z+1-i)(z+3+1)=0 \Rightarrow z_1=-1+i;\ \ \ z_2=-3-i[/tex]
[tex]\it c)\ z^2+(3-2i)z-1-3i=0 \Rightarrow z^2+3z-2iz-1-3i=0 \Rightarrow \\ \\ \Rightarrow (z^2-2iz-1)+3z-3i=0 \Rightarrow (z^2-2iz-1)+3(z-i)=0 \Rightarrow \\ \\ \Rightarrow(z^2-2iz+i^2)+3(z-i)=0 \Rightarrow (z^2-i)^2+3(z-i)=0 \Rightarrow \\ \\ \Rightarrow (z-i)(z-i+3)=0 \Rightarrow z_1=i;\ \ \ z_2=-3+i[/tex]
