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Abdellahius
@Abdellahius
April 2019
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Salut g besoin d'aide sur la première question de l'exo 94 svp!!
g commencé mais je ne c plus continué...:
z'=
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anylor
Bonsoir,
Z= x+iy
on le remplace dans l'expression de z'
z' = [( x+iy ) -2+i ] / [(x+iy) +i]
= [( x-2) +i(y +1)] / [x+ i (y+1)]
on multiplie par la quantité conjuguée
= [( x-2) +i(y +1)] / [x+ i (y+1)] × [x- i (y+1)] / [x- i (y+1)]
[x(x-2) - i(x-2)(y+1) +xi(y +1)- i²(y +1)²] / [x² -i² (y+1)²]
mais comme i² = -1
on a :
[x(x-2) +(y +1)² - i(x-2)(y+1) +xi(y +1)] / [x² +(y+1)²]
[x(x-2) +(y +1)² - i ( xy + x- 2y-2 -xy -x)) ] / [x² +(y+1)²]
(attention au signe quand on factorise par i)
(x²-2x+y²+2y+1) +2 i(y+1 ) ] / [x² +(y+1)²]
[(x²-2x+y²+2y+1)] / [x² +(y+1)²] + [ i (2y+2 )] / [x² +(y+1)²]
partie réelle de z' => [(x²-2x+y²+2y+1)] / [x² +(y+1)²]
(on retrouve la valeur de l'énoncé)
et partie imaginaire de z' => [ (2 y+2) ] / [x² +(y+1)²]
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Lista de comentários
Z= x+iy
on le remplace dans l'expression de z'
z' = [( x+iy ) -2+i ] / [(x+iy) +i]
= [( x-2) +i(y +1)] / [x+ i (y+1)]
on multiplie par la quantité conjuguée
= [( x-2) +i(y +1)] / [x+ i (y+1)] × [x- i (y+1)] / [x- i (y+1)]
[x(x-2) - i(x-2)(y+1) +xi(y +1)- i²(y +1)²] / [x² -i² (y+1)²]
mais comme i² = -1
on a :
[x(x-2) +(y +1)² - i(x-2)(y+1) +xi(y +1)] / [x² +(y+1)²]
[x(x-2) +(y +1)² - i ( xy + x- 2y-2 -xy -x)) ] / [x² +(y+1)²]
(attention au signe quand on factorise par i)
(x²-2x+y²+2y+1) +2 i(y+1 ) ] / [x² +(y+1)²]
[(x²-2x+y²+2y+1)] / [x² +(y+1)²] + [ i (2y+2 )] / [x² +(y+1)²]
partie réelle de z' => [(x²-2x+y²+2y+1)] / [x² +(y+1)²]
(on retrouve la valeur de l'énoncé)
et partie imaginaire de z' => [ (2 y+2) ] / [x² +(y+1)²]