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z-C-z-3i-z-i-R-z-3-z-1-I-find-z-

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Question Number 176513 by CrispyXYZ last updated on 20/Sep/22
z∈C, ((z−3i)/(z+i))∈R^− , ((z−3)/(z+1))∈I find z.
z∈C,z−3iz+i∈R−,z−3z+1∈Ifindz.
Answered by a.lgnaoui last updated on 20/Sep/22
Posons z=x+iy ((z−3i)/(z+i))=((x+(y−3)i)/(x+(y+1)i))=(([x+(y−3)i][x−(y+1)i])/(x^2 +(y+1)^2 ))=(([x^2 +(y+1)(y−3)]+[(x(y−3)−x(y+1))i])/(x^2 −(y+1)^2 )) ((z−3i)/(z+i))= (((x^2 +y^2 −2y−3)−(4x)i)/(x^2 +(y+1)^2 )) ((z−3)/(z+1))=(((x−3)+yi)/((x+1)+yi))=(([(x−3)+yi][(x+1)−yi])/((x+1)^2 +y^2 ))=(([(x−3)(x+1)+y^2 ]+[((x+1)y−(x−3)y)i)/((x+1)^2 +y^2 )) ((z−3)/(z+1))=(([x^2 +y^2 −2x−3]+(4y)i)/((x+1)^2 +y^2 )) ((z−3i)/(z+i))∈R^− ⇒x=0 ((z−3i)/(z+i))=((y^2 −2y−3)/((y+1)^2 ))<0 ((z−3)/(z+1)) imaginaire x^2 +y^2 −2x−3=0 (y≠0) avec x=0 donc z verifie y=±(√3) y=−(√3) rejete y=+(√3) ((z−3i)/(z+i))=((3−2(√3) −3)/(((√3) +1)^2 ))<0 et ((z−3)/(z+1))=((4(√3) )/3)i ∈I ⇒ z=(√3) i
Posonsz=x+iyz−3iz+i=x+(y−3)ix+(y+1)i=[x+(y−3)i][x−(y+1)i]x2+(y+1)2=[x2+(y+1)(y−3)]+[(x(y−3)−x(y+1))i]x2−(y+1)2z−3iz+i=(x2+y2−2y−3)−(4x)ix2+(y+1)2z−3z+1=(x−3)+yi(x+1)+yi=[(x−3)+yi][(x+1)−yi](x+1)2+y2=[(x−3)(x+1)+y2]+[((x+1)y−(x−3)y)i(x+1)2+y2z−3z+1=[x2+y2−2x−3]+(4y)i(x+1)2+y2z−3iz+i∈R−⇒x=0z−3iz+i=y2−2y−3(y+1)2<0z−3z+1imaginairex2+y2−2x−3=0(y≠0)avecx=0donczverifiey=±3y=−3rejetey=+3z−3iz+i=3−23−3(3+1)2<0etz−3z+1=433i∈I⇒z=3i

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