Question

The locus of z satisfying the inequality $lo{g}_{\frac{1}{3}}$|z+1| > $lo{g}_{\frac{1}{3}}$|z-1| is

• A. R(z)<0
• B. R(z)>0
• C. I(z)<0
• D. None of these

Answer 1

The correct option is A

R(z)<0

We know that $lo{g}_a$m>$lo{g}_a$n $\Rightarrow$ m>n or m<n, according as a>1 or 0<a<1

Hence for z=x+iy

$lo{g}_{\frac{1}{3}}$|z-1| $\Rightarrow$|z+1|<|z-1|

$$\left\{\begin{array}{c}o<\frac{1}{3}<1\end{array}\right\}$$

$\Rightarrow$ |x+iy+1|<|x+iy-1|

$\Rightarrow$ $(x+1{)}^2$ + $y^2$<$(x-1{)}^2$ + $y^2$

$\Rightarrow$ 4x<0 $\Rightarrow$ x<0 $\Rightarrow$ Re(z)<0