Question

The locus of z which satisfies the inequality log_0.3|z – 1| > log_0.3 |z – i| is given by-

• A. x + y < 0
• B. x + y > 0
• C. x – y > 0
• D. x – y < 0

Answer 1

Answer: (C)

x – y > 0

Answer 2

Sol. log_0.3 |z – 1| > log_0.3 |z – i| Ž |z – 1| < |z – i|

[\ base < 1]

Ž |z – 1|² < |z – i|² Ž (z – 1)$(\bar{z}–1)$ < (z – i) $(\bar{z} + i)$

Ž z$\bar{z}$– z – $\bar{z}$+1 < z$\bar{z}$ + iz – i$\bar{z}$ + 1 Ž (1 + i)z

+ (1 – i) $\bar{z}$> 0

Ž (z + $\bar{z}$) + i (z – $\bar{z}$) > 0 Ž $\left( \frac{z + \bar{z}}{2} \right)$

– $\left( \frac{z–\bar{z}}{2i} \right)$ > 0 Ž x – y > 0.