The region represented by the inequality ` 2 lt | z + i| le 3 ` is . . . .

Answer

To solve the inequality \( 2 < |z + i| \leq 3 \), we will break it down into two parts and analyze each part step by step.

### Step 1: Rewrite the Complex Number
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. The expression \( z + i \) can be rewritten as:
\[ z + i = x + iy + i = x + i(y + 1) \]

### Step 2: Express the Modulus
The modulus \( |z + i| \) is given by:
\[ |z + i| = |x + i(y + 1)| = \sqrt{x^2 + (y + 1)^2} \]

### Step 3: Analyze the First Part of the Inequality
The first part of the inequality is:
\[ 2 < |z + i| \]
This implies:
\[ 2 < \sqrt{x^2 + (y + 1)^2} \]
Squaring both sides gives:
\[ 4 < x^2 + (y + 1)^2 \]
This can be rewritten as:
\[ x^2 + (y + 1)^2 > 4 \]
This represents the region outside a circle centered at \( (0, -1) \) with a radius of 2.

### Step 4: Analyze the Second Part of the Inequality
The second part of the inequality is:
\[ |z + i| \leq 3 \]
This implies:
\[ \sqrt{x^2 + (y + 1)^2} \leq 3 \]
Squaring both sides gives:
\[ x^2 + (y + 1)^2 \leq 9 \]
This represents the region inside or on the boundary of a circle centered at \( (0, -1) \) with a radius of 3.

### Step 5: Combine the Results
The combined inequalities are:
\[ 4 < x^2 + (y + 1)^2 \leq 9 \]
This describes the annular region (ring-shaped area) between the two circles:
- The outer circle has a radius of 3.
- The inner circle has a radius of 2.

### Conclusion
Thus, the region represented by the inequality \( 2 < |z + i| \leq 3 \) is the area between the two circles centered at \( (0, -1) \) with radii 2 and 3, where the boundary of the outer circle is included, and the boundary of the inner circle is excluded.

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