Question

The locus of 2 satisfying the inequality
\[{\log _{\dfrac{1}{3}}}\left| {z + 1} \right| > {\log _{\dfrac{1}{3}}}\left| {z - 1} \right|\]is
A.\[R(z) < 0\]
B.\[R(z) > 0\]
C.\[I(z) < 0\]
D.None of these

Answer 1

Hint: Use the property =>\[{\log _a}b > {\log _a}c\] ; where \[a < 1\]
                                                          \[ = b < c\]
Then proceed by solving the \[\left| {z + 1} \right| < \left| {z - 1} \right|\]
By substituting \[z = x + iy\]
Given: an equation: \[{\log _{\dfrac{1}{3}}}\left| {z + 1} \right| > {\log _{\dfrac{1}{3}}}\left| {z - 1} \right|\]

Complete step by step solution:
Step 1: In this question, we are given
\[{\log _{\dfrac{1}{3}}}\left| {z + 1} \right| > {\log _{\dfrac{1}{3}}}\left| {z - 1} \right|\]
As we know that,
\[{\log _a}b > {\log _a}c\] ; if \[a > 1\]
 Then, \[b > c\]
Whereas if\[a < 1\],
We get \[{\log _a}b > {\log _a}c\]
\[b < c\].
Step 2:
Applying this in given equation, we get,
\[{\log _{\dfrac{1}{3}}}\left| {z + 1} \right| > {\log _{\dfrac{1}{3}}}\left| {z - 1} \right|\]\[\because \dfrac{1}{3} < 1\]
So, \[\left| {z + 1} \right| < \left| {z - 1} \right|\]
Step 3: Since, we know a complex number z is equal to,
\[z = x + iy\]
Substituting this in above equation we get,
\[\left| {(x + 1) + iy} \right| < \left| {(x - 1) + iy} \right|\]
Now, squaring both sides, so as to ensure a positive value, we get,
\[{\left| {(x + 1) + (iy)} \right|^2} < {\left| {(x - 1) + (iy)} \right|^2}\]
We get,
\[{(x + 1)^2} + {(iy)^2} < {(x - 1)^2} + {(iy)^2}\]
We get, a relation,
\[{(x + 1)^2} < {(x - 1)^2}\] ………….(1)
Step 4:
On solving the relation (1),
We get,
\[{x^2} + 1 + 2x < {x^2} + 1 - 2x\]
Or, \[4x < 0\]
Or, \[x < 0\].
Since here \[x\] refers to the real part of z,
We get,
\[R(z) < 0\]
Hence the correct answer is (A).

Note: In this question, it is important to remember the logarithmic relation between two variables. Since if we proceed with the relation, \[{\log _a}b > {\log _a}c\] ; \[a > 1\]. We get, \[b < c\], as a result we end up with \[R(z) > 0\]; which is completely wrong.