The region represented by the inequality |2z-3i|<|3z-2i| is

Options

A. the unit disc with its centre at `z=0`
B. the exterior of the unit circle with its centre at `z=0`
C. the inerior of a square of side `2` units with its centre at `z=0`
D. none of these

Explanation

To solve the inequality |2z - 3i| < |3z - 2i|, we will follow these steps:

### Step 1: Substitute z with x + iy
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

### Step 2: Rewrite the inequality
Substituting \( z \) into the inequality gives:
\[ |2(x + iy) - 3i| < |3(x + iy) - 2i| \]
This simplifies to:
\[ |2x + 2iy - 3i| < |3x + 3iy - 2i| \]
Which can be rewritten as:
\[ |2x + i(2y - 3)| < |3x + i(3y - 2)| \]

### Step 3: Express the moduli
Now we express the moduli:
\[ \sqrt{(2x)^2 + (2y - 3)^2} < \sqrt{(3x)^2 + (3y - 2)^2} \]

### Step 4: Square both sides
To eliminate the square roots, we square both sides:
\[ (2x)^2 + (2y - 3)^2 < (3x)^2 + (3y - 2)^2 \]
Expanding both sides:
\[ 4x^2 + (4y^2 - 12y + 9) < 9x^2 + (9y^2 - 12y + 4) \]

### Step 5: Rearranging the inequality
Rearranging gives:
\[ 4x^2 + 4y^2 - 12y + 9 < 9x^2 + 9y^2 - 12y + 4 \]
Simplifying this results in:
\[ 4x^2 + 4y^2 + 9 < 9x^2 + 9y^2 + 4 \]
This can be rearranged to:
\[ 0 < 5x^2 + 5y^2 - 5 \]
Dividing through by 5 gives:
\[ 0 < x^2 + y^2 - 1 \]
Thus:
\[ x^2 + y^2 > 1 \]

### Step 6: Interpret the result
The inequality \( x^2 + y^2 > 1 \) represents the region outside the circle of radius 1 centered at the origin in the complex plane.

### Conclusion
The region represented by the inequality |2z - 3i| < |3z - 2i| is the exterior of the unit circle centered at the origin.

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