If `|z - 4| lt |z - 2|`, then

Options

A. `Re (z) gt 0`
B. `Re (z) lt 0`
C. `Re (z) gt 3`
D. `Re (z) gt 2`

Answer

To solve the inequality \( |z - 4| < |z - 2| \), we will express \( z \) in terms of its real and imaginary parts. Let's denote \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part of \( z \).

### Step-by-Step Solution:

1. **Express the Moduli**:
We start with the given inequality:
\[ |z - 4| < |z - 2| \]
Substituting \( z = x + iy \), we have:
\[ |(x + iy) - 4| < |(x + iy) - 2| \]
This simplifies to:
\[ |(x - 4) + iy| < |(x - 2) + iy| \]

2. **Calculate the Moduli**:
The modulus of a complex number \( a + ib \) is given by \( \sqrt{a^2 + b^2} \). Therefore, we can express the moduli as:
\[ \sqrt{(x - 4)^2 + y^2} < \sqrt{(x - 2)^2 + y^2} \]

3. **Square Both Sides**:
Since both sides are positive, we can square both sides without changing the inequality:
\[ (x - 4)^2 + y^2 < (x - 2)^2 + y^2 \]

4. **Cancel \( y^2 \)**:
We can cancel \( y^2 \) from both sides:
\[ (x - 4)^2 < (x - 2)^2 \]

5. **Expand Both Sides**:
Expanding both sides gives:
\[ (x^2 - 8x + 16) < (x^2 - 4x + 4) \]

6. **Simplify the Inequality**:
Subtract \( x^2 \) from both sides:
\[ -8x + 16 < -4x + 4 \]
Rearranging this gives:
\[ -8x + 4x < 4 - 16 \]
Which simplifies to:
\[ -4x < -12 \]

7. **Divide by -4**:
When dividing by a negative number, we reverse the inequality:
\[ x > 3 \]

8. **Conclusion**:
Since \( x \) is the real part of \( z \), we can conclude:
\[ \text{Re}(z) > 3 \]

### Final Answer:
Thus, the condition on the real part of \( z \) is:
\[ \text{Re}(z) > 3 \]