If zx +iyand show that |w 1, z is purely real

1.	If zx +iyand show that \|w 1, z is purely real
Answer» z=x+iy in w gives w=[(1+y)-ix]/[(x-1)+iy] on solving gives: w=[(1+y)-ix]*[(x-1)+iy] / [(x-1)^2+y^2] = {(x-y-1) -i(x(x-1)+y(y+1))} / [(x-1)^2+y^2] as \|w\|=1 , so {(x-y-1)^2+ (x(x-1)+y(y+1))^2} = [(x-1)^2+y^2]^2 on solving the above equation, we have 2x^3+2y^3+2xy(x+y)-2x^2-2xy+2(x+y) = 0 which is (x+y).(x^2+y^2-2x+2) =0 is the locus of Z

Answer»

z=x+iy in w gives w=[(1+y)-ix]/[(x-1)+iy]

on solving gives: w=[(1+y)-ix]*[(x-1)+iy] / [(x-1)^2+y^2]

= {(x-y-1) -i(x(x-1)+y(y+1))} / [(x-1)^2+y^2]

as |w|=1 , so {(x-y-1)^2+ (x(x-1)+y(y+1))^2} = [(x-1)^2+y^2]^2

on solving the above equation, we have

2x^3+2y^3+2xy(x+y)-2x^2-2xy+2(x+y) = 0

which is (x+y).(x^2+y^2-2x+2) =0 is the locus of Z

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