If ` log _(1//3) | z + 1| gt log _(1//3) | z - 1| ` : then

Options

A. `Re z lt 0 `
B. ` Re z gt 0 `
C. `Re z ge 0 `
D. none

Answer

To solve the inequality \( \log_{(1/3)} |z + 1| > \log_{(1/3)} |z - 1| \), we can follow these steps:

### Step 1: Understand the logarithmic inequality
Given the properties of logarithms, if \( \log_b a > \log_b c \) and \( 0 < b < 1 \), then it implies \( a < c \). Here, our base \( \frac{1}{3} \) is less than 1.

### Step 2: Rewrite the inequality
From the inequality \( \log_{(1/3)} |z + 1| > \log_{(1/3)} |z - 1| \), we can rewrite it as:
\[ |z + 1| < |z - 1| \]

### Step 3: Express \( z \) in terms of its real and imaginary parts
Let \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part. Then we have:
\[ |z + 1| = |(x + 1) + iy| = \sqrt{(x + 1)^2 + y^2} \]
\[ |z - 1| = |(x - 1) + iy| = \sqrt{(x - 1)^2 + y^2} \]

### Step 4: Set up the inequality
Now we can set up the inequality:
\[ \sqrt{(x + 1)^2 + y^2} < \sqrt{(x - 1)^2 + y^2} \]

### Step 5: Square both sides
Squaring both sides (since both sides are positive), we get:
\[ (x + 1)^2 + y^2 < (x - 1)^2 + y^2 \]

### Step 6: Simplify the inequality
The \( y^2 \) terms cancel out:
\[ (x + 1)^2 < (x - 1)^2 \]
Expanding both sides:
\[ x^2 + 2x + 1 < x^2 - 2x + 1 \]
Now, simplify:
\[ 2x < -2x \]
\[ 4x < 0 \]
Thus, we find:
\[ x < 0 \]

### Conclusion
The real part of \( z \) is less than 0:
\[ \text{Re}(z) < 0 \]