Question

Let $z(\ne 2)$ be a complex number such that ${\log }_{1/2}|z-2|>{\log }_{1/2}|z|$, then

• A. $Re\left(z\right)>1$
• B. $Im\left(z\right)>1$
• C. $Re\left(z\right)=1$
• D. $Im\left(z\right)=1$

Answer 1

The correct option is A $Re\left(z\right)>1$

Let $z=x+iy$
${\log }_{1/2}|z-2|>{\log }_{1/2}\left|z\right|$
Therefore, ${\log }_{1/2}|x-2+iy|>{\log }_{1/2}|x+iy|$
${\log }_{1/2}\sqrt{{(x-2)}^2+y^2}>{\log }_{1/2}\sqrt{x^2+y^2}$
$\frac{1}{2}{\log }_{1/2}[{(x-2)}^2+y^2]>\frac{1}{2}{\log }_{1/2}(x^2+y^2)$
$\frac{1}{2}[{(x-2)}^2+y^2]-{\log }_{1/2}(x^2+y^2)>0$
${\log }_{1/2}\frac{{(x-2)}^2+y^2}{x^2+y^2}>0$
${}_{\left[\frac{1}{2}\right]}{\log }_{1/2}\frac{{(x-2)}^2+y^2}{x^2+y^2}<{\left[\frac{1}{2}\right]}^0$
[when base $<1$, inequality reverses]
Thus $\frac{{(x-2)}^2+y^2}{x^2+y^2}<1$
$\Rightarrow {(x-2)}^2+y^2<x^2+y^2$
$\Rightarrow {(x-2)}^2<x^2$
Also $|x-2|<\left|x\right|$
$\Rightarrow -x<x-2<x$
$\Rightarrow x-2>-x$
$\Rightarrow 2x>2$
$\Rightarrow x>1$
Hence, $Re\left(z\right)>1$.