[tex]M = (x;y),\ \ \ z = x + i y,\ \ \ \bar{z} = x - i y\\ M_1 = (x_1 ; y_1),\ \ \ z_1 = (1 + i)\bar{z}-2=(1+i)(x-i\ y)-2\\z_1=x-iy+i\ x-i^2y=(x+y-2)+i\ (x-y)\\M_1=(x_1;y_1)=(x+y-2 ; x-y) \\\\M_2=(x_2;y_2),\ \ \ \ z_2=i\ \bar{z}+2+i=i(x-iy)+2+i\\z_2=ix-i^2y+2+i=(2+y)+i(x+1)\\M_2=(x_2;y_2)=(y+2;x+1)\\
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1)
[tex]le\ pente\ du\ droit\ OM_2=\frac{y_2-0}{x_2-0}=\frac{x+1}{y+2}\\\\le\ pente\ du\ droit\ OM_1=\frac{y_1-0}{x_1-0}=\frac{x-y}{x+y-2}\\\\Si\ et\ seulement\ si\ O,M_1\ et\ M_2\ sont\ alignes,\ les\ pentes\ sont\ egales.\\\\Donc,\ \frac{x+1}{y+2}=\frac{x-y}{x+y-2}\\\\x^2+xy-2x+x+y-2=xy+2x-y^2-2y\\\\x^2+y^2-3x+3y-2=0\\[/tex]
2)
[tex]a=\frac{3}{2}(1 -i),\ \ \ A=(\frac{3}{2};-\frac{3}{2})\\\\M=(x;y),\ \ \ z=x+iy\\\\AM^2=\frac{13}{2}\\\\(x-\frac{3}{2})^2+(y+\frac{3}{2})^2=\frac{13}{2},\ \ \ denote\ un\ cercle\\\\x^2-3x+\frac{9}{4}+y^2+3y+\frac{9}{4}=\frac{13}{2}\\\\x^2+y^2-3x+3y=2\\[/tex]
Les points M(x;y) doit appartenir à l'ensemble ε des points tel que :
x² + y² -3x + 3y = 2
La nature du ensemble ε : un cercle du centre A(3/2,-3/2) et du rayon √13/√2.
