determiner z tel que z=z^2 −z+2


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Question Number 176090 by doline last updated on 11/Sep/22

$d e t e r m i n e r z t e l q u e z = z^{2} - z + 2$
Commented by Frix last updated on 12/Sep/22

$why not simply use$ $z^{2} + p z + q = 0 \Rightarrow z = - \frac{p}{2} \pm \sqrt{\frac{p^{2}}{4} - q}$ $? ? ?$
	Answered by a.lgnaoui last updated on 12/Sep/22

	$z^{2} - z - z + 2 = 0$ $(z^{2} - 2 z + 1) + 1 = 0$ ${(z - 1)}^{2} = - 1 = i^{2}$ $(z - 1 - i) (z - 1 + i) = 0$ $s o l u t i o n s :$ $z = 1 + i$ $z = 1 - i$
	Answered by Rasheed.Sindhi last updated on 13/Sep/22
	$z^2 −2z+2=0 (x+iy)^2 −2(x+iy)+2=0; x,y∈R x^2 −y^2 +2ixy−2x−2iy+2=0 { ((x^2 −y^2 −2x+2=0)),((2xy−2y=0⇒2y(x−1)=0⇒y(x−1)=0)) :} y(x−1)=0=0⇒y=0∨x=1 y=0: x^2 −y^2 −2x+2=0 x^2 −2x+2 x=((2±(√(4−8)))/2)=1±i∉R (Rejected) x=1: 1−y^2 −2+2=0 y=±1 z=x+iy=1±i$
	$z^{2} - 2 z + 2 = 0$ ${(x + i y)}^{2} - 2 (x + i y) + 2 = 0; x, y \in R$ $x^{2} - y^{2} + 2 i x y - 2 x - 2 i y + 2 = 0$ ${\begin{cases} x^{2} - y^{2} - 2 x + 2 = 0 \\ 2 x y - 2 y = 0 \Rightarrow 2 y (x - 1) = 0 \Rightarrow y (x - 1) = 0 \end{cases}$ $y (x - 1) = 0 = 0 \Rightarrow y = 0 \lor x = 1$ $y = 0 :$ $x^{2} - y^{2} - 2 x + 2 = 0$ $x^{2} - 2 x + 2$ $x = \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i \notin R (Rejected)$ $x = 1 :$ $1 - y^{2} - 2 + 2 = 0$ $y = \pm 1$ $z = x + i y = 1 \pm i$
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