Question

Aflati z complex astfel incat :
z^{2} +z = -9+3i

Answer 1

[tex]z^2+z=-9+3i \\ \\ Inlocuim \\ \\ z=x+yi \\ \\ \left(x+yi\right)^2+x+yi=-9+3i \\ \\ \left(x^2-y^2+x\right)+i\left(2xy+y\right)=-9+3i \\ \\ \begin{bmatrix}x^2-y^2+x=-9\\ 2yx+y=3\end{bmatrix} \\ \\ 2yx+y=3 \\ \\ 2yx+y-y=3-y \\ \\ 2yx=3-y \\ \\ \frac{2yx}{2y}=\frac{3}{2y}-\frac{y}{2y} \\ \\ x=\frac{3-y}{2y} \\ \\ \left(\frac{3-y}{2y}\right)^2-y^2+\frac{3-y}{2y}=-9 \\ \\ \left(\frac{3-y}{2y}\right)^2\cdot \:2y-y^2\cdot \:2y+\frac{3-y}{2y}\cdot \:2y=-9\cdot \:2y \\ \\ [/tex][tex]\frac{\left(3-y\right)^2}{2y}-2y^3+3-y=-18y \\ \\ -2y^3-\frac{y}{2}-3+\frac{9}{2y}+3=-18y \\ \\ 2y^3\cdot \:2y-\frac{y}{2}\cdot \:2y-3\cdot \:2y+\frac{9}{2y}\cdot \:2y+3\cdot \:2y=-18y\cdot \:2y \\ \\ -4y^4-y^2-6y+9+6y=-36y^2 \\ \\ -4y^4-y^2+9=-36y^2 \\ \\ 4y^4-y^2+9+36y^2=-36y^2+36y^2 \\ \\ -4y^4+35y^2+9=0 \\ \\ u=y^2 si u^2=y^4 \\ \\ -4u^2+35u+9=0 \\ \\ u=\frac{-35+\sqrt{35^2-4\left(-4\right)9}}{2\left(-4\right)}:\quad -\frac{1}{4} \\ \\ [/tex][tex]u=\frac{-35-\sqrt{35^2-4\left(-4\right)9}}{2\left(-4\right)}:\quad 9 \\ \\ u=-\frac{1}{4},\:u=9 \\ \\ y^2=-\frac{1}{4} \\ \\ y^2=9 \\ \\ y=\sqrt{9},\:y=-\sqrt{9} \\ \\ y=3,\:y=-3 \\ \\ 2\cdot \:3x+3=3 \\ \\ 2\cdot \:3x+3-3=3-3 \\ \\ 2\cdot \:3x=0 \\ \\ \frac{2\cdot \:3x}{6}=\frac{0}{6} \\ \\ x=0 \\ \\ 2\left(-3\right)x-3=3 \\ \\ -2\cdot \:3x-3=3 \\ \\ -6x-3=3 \\ \\ -6x-3+3=3+3 \\ \\ -6x=6 \\ \\ x=-1 \\ \\ \begin{pmatrix}y=3,\:&x=0\\ y=-3,\:&x=-1\end{pmatrix} \\ \\ [/tex]$z=3i,\:z=-1-3i$