Solveeit Logo
Login

Question

Question: The locus of z (= x + iy) which satisfies the inequality log<sub>0.3</sub> \|z – 1\| \> log<sub>0.3<...

The locus of z (= x + iy) which satisfies the inequality log0.3 |z – 1| > log0.3 |z – i| is given by –

A

x + y < 0

B

x – y > 0

C

x + y > 0

D

x – y < 0

Answer

x – y > 0

Explanation

Solution

Sol. We have log0.3 |z – 1| > log0.3 |z –i|

Ž |z – 1| < |z – i|(Q 0 < 0.3 < 1)

Ž |x + iy – 1| < |x + iy – i|

Ž (x – 1)2 + y2 < x2 + (y – 1)2

Ž x2 + y2 – 2x + 1 < x2 + y2 – 2y + 1

Ž –2(x – y) < 0 Ž x – y > 0 .