The inequality `|z-i| lt |z + i|` represents the region

Options

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

Answer

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

### Step-by-Step Solution:

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 of \( z \).

2. **Rewrite the inequality**:
The inequality becomes:
\[ |(x + iy) - i| < |(x + iy) + i| \]
This simplifies to:
\[ |x + i(y - 1)| < |x + i(y + 1)| \]

3. **Calculate the moduli**:
The modulus \( |x + i(y - 1)| \) is given by:
\[ \sqrt{x^2 + (y - 1)^2} \]
The modulus \( |x + i(y + 1)| \) is given by:
\[ \sqrt{x^2 + (y + 1)^2} \]

4. **Set up the inequality**:
The inequality now reads:
\[ \sqrt{x^2 + (y - 1)^2} < \sqrt{x^2 + (y + 1)^2} \]

5. **Square both sides**:
To eliminate the square roots, we square both sides:
\[ x^2 + (y - 1)^2 < x^2 + (y + 1)^2 \]

6. **Simplify the inequality**:
Cancel \( x^2 \) from both sides:
\[ (y - 1)^2 < (y + 1)^2 \]
Expanding both sides gives:
\[ y^2 - 2y + 1 < y^2 + 2y + 1 \]

7. **Further simplification**:
Cancel \( y^2 + 1 \) from both sides:
\[ -2y < 2y \]
Rearranging gives:
\[ -4y < 0 \]

8. **Final result**:
Dividing by -4 (and reversing the inequality) results in:
\[ y > 0 \]

### Conclusion:
The inequality \( |z - i| < |z + i| \) represents the region where the imaginary part of \( z \) is positive, i.e., the upper half of the complex plane.