The inequality `|z+2| lt |z-2|` represents the region given by

Options

A. `Re(z)gt0`
B. `Re(z)lt0`
C. `Re(z)gt2`
D. none of these

Answer

To solve the inequality \( |z + 2| < |z - 2| \), 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 of the complex number \( z \).

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

### Step 3: Square both sides to eliminate the square roots
Since both sides are positive, we can square both sides without changing the inequality:
\[ (x + 2)^2 + y^2 < (x - 2)^2 + y^2 \]

### Step 4: Simplify the inequality
Now, we can simplify the inequality:
\[ (x + 2)^2 < (x - 2)^2 \]
Expanding both sides:
\[ x^2 + 4x + 4 < x^2 - 4x + 4 \]

### Step 5: Cancel out the common terms
We can cancel \( x^2 \) and \( 4 \) from both sides:
\[ 4x < -4x \]

### Step 6: Combine like terms
Adding \( 4x \) to both sides gives:
\[ 8x < 0 \]

### Step 7: Solve for \( x \)
Dividing both sides by 8 (which does not change the direction of the inequality):
\[ x < 0 \]

### Conclusion
The inequality \( |z + 2| < |z - 2| \) represents the region where the real part of \( z \) is less than 0. Thus, the region is the left half of the complex plane.