If `z in C - {0, -2}` is such that `log_((1//7)) |z-2| gt log_((1//7)) |z|` then

Options

A. `Re (z) gt 1`
B. `Re (z) lt 1`
C. `Im (z) gt 1`
D. `Im (z) lt 1`

Answer

To solve the inequality given in the problem, we will follow these steps:

### Step 1: Understand the given inequality
We are given that:
\[ \log_{(1/7)} |z - 2| > \log_{(1/7)} |z| \]
Since the base \( \frac{1}{7} \) is less than 1, we know that the logarithmic function is decreasing. Therefore, we can rewrite the inequality as:
\[ |z - 2| < |z| \]

### Step 2: 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 of \( z \). We can then express the moduli:
\[ |z| = \sqrt{x^2 + y^2} \]
\[ |z - 2| = |(x - 2) + iy| = \sqrt{(x - 2)^2 + y^2} \]

### Step 3: Substitute the expressions into the inequality
Substituting the expressions for the moduli into the inequality gives:
\[ \sqrt{(x - 2)^2 + y^2} < \sqrt{x^2 + y^2} \]

### Step 4: Square both sides of the inequality
To eliminate the square roots, we square both sides:
\[ (x - 2)^2 + y^2 < x^2 + y^2 \]

### Step 5: Simplify the inequality
Expanding the left side:
\[ x^2 - 4x + 4 + y^2 < x^2 + y^2 \]
Now, we can cancel \( x^2 \) and \( y^2 \) from both sides:
\[ -4x + 4 < 0 \]

### Step 6: Solve for \( x \)
Rearranging the inequality gives:
\[ 4 < 4x \]
Dividing both sides by 4:
\[ 1 < x \]
or
\[ x > 1 \]

### Conclusion
The solution to the inequality is that the real part of \( z \) must be greater than 1. Therefore, the values of \( z \) must satisfy:
\[ \text{Re}(z) > 1 \]