Question

Locate the complex numbers $z=x+iy$ for which ${\log }_{1/2}|z-2|>{\log }_{1/2}\left|z\right|$

• A. ${\log }_{1/2}|z-2|>{\log }_{1/2}$ z represents all points to the right of line x=3/2 except (3,0) represented by z=3.
• B. ${\log }_{1/2}|z-2|>{\log }_{1/2}$ z represents all points to the right of line x=1, except (2,0) represented by z=2.
• C. ${\log }_{1/2}|z-2|>{\log }_{1/2}$ z represents all points to the left of line x=1 except (3,0) represented by z=3.
• D. ${\log }_{1/2}|z-3|>{\log }_{1/2}$ z represents all points to the left of line x=3/2 except (2,0) represented by z=3.

Answer 1

The correct option is B ${\log }_{1/2}|z-2|>{\log }_{1/2}$ z represents all points to the right of line x=1, except (2,0) represented by z=2.

On simplifying given equation, we get

${\log }_{0.5}|z-2|>{\log }_{0.5}|z|$

$-{\log }_2|z-2|>-{\log }_2|z|$

${\log }_2|z-2|<{\log }_2|z|$

$|z-2|<|z|$

$(x-2{)}^2+y^2<x^2+y^2$ ...considering $z=x+iy$

Hence, $(2x-2)\left(2\right)>0$

$x-1>0$

$x>1$

Hence it represents all the points on the right of the line $x=1$.

But, $z\ne 2$ since inequality doesnt hold at $z=2.$