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yunusemreeee
@yunusemreeee
January 2021
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Je suis bloqué sur ces exercices, aidez moi s'il vous plait
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scoladan
Verified answer
Bonjour,
III) z = a + ib
==> Z = [(a-2) + ib]/[a + (b+2)i]
= [(a-2) + ib]x[a - (b+2)i] /[a^2 + (b+2)^2]
= [ a(a-2) + b(b+2) + ( -(a-2)(b+2) + ab )i ] / [a^2 + (b+2)^2]
a) Z réel ==> -(a-2)(b+2) + ab = 0
<=> -(ab + 2a -2b - 4) + ab = 0
<=> -2a + 2b = -4
<=> (a-b) = 2
<=> b = a-2
L'ensemble des points M est la droite d'équation y = x-2
b) Z imaginaire pur
==> a(a-2) + b(b+2) = 0
<=> a^2 - 2a + b^2 + 2b = 0
<=> (a-1)^2 -1 + (b+1)^2 -1 = 0
<=> (a-1)^2 + (b+1)^2 = 2
==> Cercle de centre P(1;-1) et de rayon racine(2)
IV)
1) Z = (1-2i)(a+ib) - (a-ib)
= a + ib - 2ia + 2b - a + ib
= (2b) + 2(b -a)i
2) Z réel ==> b-a = 0 ==> b=a ==> Droite d'équation y = x
3) Z imaginaire ==> b=0 ==> Axe des réels y = 0
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Verified answer
Bonjour,III) z = a + ib
==> Z = [(a-2) + ib]/[a + (b+2)i]
= [(a-2) + ib]x[a - (b+2)i] /[a^2 + (b+2)^2]
= [ a(a-2) + b(b+2) + ( -(a-2)(b+2) + ab )i ] / [a^2 + (b+2)^2]
a) Z réel ==> -(a-2)(b+2) + ab = 0
<=> -(ab + 2a -2b - 4) + ab = 0
<=> -2a + 2b = -4
<=> (a-b) = 2
<=> b = a-2
L'ensemble des points M est la droite d'équation y = x-2
b) Z imaginaire pur
==> a(a-2) + b(b+2) = 0
<=> a^2 - 2a + b^2 + 2b = 0
<=> (a-1)^2 -1 + (b+1)^2 -1 = 0
<=> (a-1)^2 + (b+1)^2 = 2
==> Cercle de centre P(1;-1) et de rayon racine(2)
IV)
1) Z = (1-2i)(a+ib) - (a-ib)
= a + ib - 2ia + 2b - a + ib
= (2b) + 2(b -a)i
2) Z réel ==> b-a = 0 ==> b=a ==> Droite d'équation y = x
3) Z imaginaire ==> b=0 ==> Axe des réels y = 0