Question

The locus of $z=x+iy$ which satisfying the inequality ${\log }_{1/2}|z-1|>{\log }_{1/2}|z-i|$ is given by

• A. $x+y<0$
• B. $x-y>0$
• C. $x-y<0$
• D. $x+y>0$

Answer 1

The correct option is B $x-y>0$

$lo{g}_{0.5}|z-1|>lo{g}_{0.5}|z-i|$
$-lo{g}_2|z-1|>-lo{g}_2|z-i|$
$lo{g}_2|z-1|<lo{g}_2|z-i|$
Hence
$|z-1|<|z-i|$
Now
Let $z=x+iy$
Hence by squaring both sides
$(x-1{)}^2+y^2<x^2+(y-1{)}^2$
$(2y-1)<(2x-1)$
$2(y-x)<0$
$y-x<0$
Or
$x-y>0$