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

Options

A. Re `(z) gt 0`
B. Re `(z) lt0`
C. Re `(z) gt 2`
D. None of these

Answer

To solve the inequality \( |z - 4| < |z - 2| \), we will follow these steps:

### Step 1: 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.

### Step 2: Rewrite the inequality using the definition of modulus
The inequality can be rewritten as:
\[ |z - 4| = |(x - 4) + iy| = \sqrt{(x - 4)^2 + y^2} \]
\[ |z - 2| = |(x - 2) + iy| = \sqrt{(x - 2)^2 + y^2} \]
Thus, the inequality becomes:
\[ \sqrt{(x - 4)^2 + y^2} < \sqrt{(x - 2)^2 + y^2} \]

### Step 3: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ (x - 4)^2 + y^2 < (x - 2)^2 + y^2 \]

### Step 4: Simplify the inequality
Since \( y^2 \) appears on both sides, we can cancel it out:
\[ (x - 4)^2 < (x - 2)^2 \]

### Step 5: Expand both sides
Expanding both sides results in:
\[ x^2 - 8x + 16 < x^2 - 4x + 4 \]

### Step 6: Rearrange the inequality
Subtract \( x^2 \) from both sides:
\[ -8x + 16 < -4x + 4 \]
Now, rearranging gives:
\[ -8x + 4x < 4 - 16 \]
\[ -4x < -12 \]

### Step 7: Divide by -4 (remember to reverse the inequality)
Dividing by -4 gives:
\[ x > 3 \]

### Step 8: Interpret the result
The inequality \( x > 3 \) indicates that the real part of \( z \) must be greater than 3. Therefore, the region represented by the inequality \( |z - 4| < |z - 2| \) is the set of all complex numbers \( z \) whose real part is greater than 3.

### Conclusion
The inequality \( |z - 4| < |z - 2| \) represents the region in the complex plane where the real part of \( z \) is greater than 3.

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