determiner-z-tel-que-z-z-2-z-2-

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} - 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