The inequality `|z-4| < |z-2|` represents

Options

A. Re(z) `>= 0`
B. Re(z) `lt 0`
C. Re(z) `gt 0`
D. None of these

Answer

To solve the inequality \( |z - 4| < |z - 2| \), we start by letting \( z \) be a complex number expressed as \( z = x + iy \), where \( x \) and \( y \) are real numbers.

### Step 1: Rewrite the inequality
We can rewrite the inequality as:
\[ |z - 4| < |z - 2| \]
Substituting \( z = x + iy \):
\[ | (x + iy) - 4 | < | (x + iy) - 2 | \]
This simplifies to:
\[ | (x - 4) + iy | < | (x - 2) + iy | \]

### Step 2: Calculate the moduli
The modulus of a complex number \( a + bi \) is given by \( \sqrt{a^2 + b^2} \). Thus, we have:
\[ | (x - 4) + iy | = \sqrt{(x - 4)^2 + y^2} \]
and
\[ | (x - 2) + iy | = \sqrt{(x - 2)^2 + y^2} \]

### Step 3: Set up the inequality
Now we can set up the inequality:
\[ \sqrt{(x - 4)^2 + y^2} < \sqrt{(x - 2)^2 + y^2} \]

### Step 4: Square both sides
To eliminate the square roots, we square both sides (noting that both sides are non-negative):
\[ (x - 4)^2 + y^2 < (x - 2)^2 + y^2 \]

### Step 5: Simplify the inequality
We can simplify this by subtracting \( y^2 \) from both sides:
\[ (x - 4)^2 < (x - 2)^2 \]
Expanding both sides:
\[ (x^2 - 8x + 16) < (x^2 - 4x + 4) \]

### Step 6: Rearranging the terms
Now, we can rearrange the terms:
\[ x^2 - 8x + 16 - x^2 + 4x - 4 < 0 \]
This simplifies to:
\[ -4x + 12 < 0 \]

### Step 7: Solve for \( x \)
Now, we solve for \( x \):
\[ -4x < -12 \implies x > 3 \]

### Step 8: Conclusion
The inequality \( |z - 4| < |z - 2| \) represents the region in the complex plane where the real part \( x \) of the complex number \( z \) is greater than 3. This means the solution represents the half-plane to the right of the vertical line \( x = 3 \).