Complexe

[suhail]

bonjour

AIDEZ Moi a faire ces exos

MERCI

NB la 1ere image est une erreur veillez m'excusez

[Barbidoux]

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(2+i)*z=1-3*i

z=(1-3*i)/(2-i)=(1-3*i)*(2+i)/((2-i)*(2+i)=(1-i)

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1+3*i*z=2-5*i

z=(1-5*i)/(3*i)=-(5+i)/3

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(3-2*i)*z+4-5*i=0

z=(-4+5*i)/(3-2*i)=(-22+7*i)/13

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7*z-4*i=-i*z+5

z*(7+i)=5+4*i

z=(5+4*i)/(7+i)=(39+23*i)/50

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(z+2)/(z-2)=1+i

(z+2-(z+z*i-2-2*i))=0

(z*i-4+2*i)/(z-2)=0

z*i-4+2*i=0 ==> z=2-4*i

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(z-3*i)/(z+2)=2-i

(z-3*i)/(z+2)-2-i=0

((z-3*i)-(2+i)*(z+2))/(z+2) =0

(-z*(1+i)-4-5*i)/(z+2) =0

(-z*(1+i)-4-5*i)=0

z=(-4-5*i)/(1+i)=-(3+5*i)/2