Question

If z = x + yi and $\frac{|z-1-i|+4}{3|z-1-i|-2}=1$, show that $x^2+y^2-2x-2y-7=0$.

Answer 1

$\frac{|z-1-i|+4}{3|z-1-i|-2}=1$

Put $z=x+yi$

$\frac{|x+iy-1-i|+4}{3|x+iy-1-i|-2}=1$

$\frac{|(x-1)+i(y-1)|+4}{3|(x-1)+i(y-1)-2}=1$

$\frac{\sqrt{(x-1{)}^2+(y-1{)}^2}+4}{3\sqrt{(x-1{)}^2+(y-1{)}^2}-2}=1$

$\sqrt{(x-1{)}^2+(y+1{)}^2+4}=3\sqrt{(x-1{)}^2+(y-1{)}^2}-2$

$2\sqrt{(x-1{)}^2+(y-1{)}^2}=6$

$\sqrt{(x-1{)}^2+(y-1{)}^2}=3$

$\sqrt{(x^2+1-2x+y^2+1-2y)}=9$

$x^2+y^2-2x-2y-7=0$ Hence proved