Question

lf ${\log }_{\left(\frac{1}{3}\right)}|z+1|>{\log }_{\left(\frac{1}{3}\right)}|z-1|$, then

• A. $Re\left(z\right)\ge 0$
• B. $Re\left(z\right)<0$
• C. $Im\left(z\right)>0$
• D. $Im\left(z\right)\le 0$

Answer 1

Answer: (B)

$Re\left(z\right)<0$

$\log \left(\frac{1}{3}\right)|z+1|>\log \left(\frac{1}{3}\right)|z-1|$
$\Rightarrow |z+1|<|z-1|$
$\Rightarrow (x^2+2x+1+y^2)<(x^2-2x+1+y^2)$
$\Rightarrow |4x|<0$
Thus, $Re\left(z\right)<0.$