The locus of z satisfying the inequality (\log\_{1/3}|z + 1... | MATH - INTEGRAL POWER OF IOTA, ALGEBRAIC OPERATIONS AND EQUALITY OF COMPLEX NUMBERS | SolveeIt

The locus of z satisfying the inequality

$\log_{1/3}|z + 1| > \log_{1/3}|z - 1|$ is

$R(z) < 0$

$R(z) > 0$

$I(z) < 0$

None

Answer

$R(z) < 0$

Explanation

Sol. $\log_{1/3}|z + 1| > \log_{1/3}|z - 1|$

⇒ $|z + 1| < |z - 1|$ ⇒ $x^{2} + 1 + 2x + y^{2} < x^{2} + 1 - 2x + y^{2}$

⇒ $x < 0$ ⇒ ${Re}(z) < 0.$

Question: The locus of z satisfying the inequality \(\log_{1/3}|z + 1| > \log_{1/3}|z - 1|\) is...