Given `z_(1)=1 + 2i`. Determine the region in the complex plane represented by `1 lt |z-z_(1)| le 3`. Represent it with the help of an Argand diagram

Answer

To solve the problem, we need to determine the region in the complex plane represented by the inequality \( 1 < |z - z_1| \leq 3 \), where \( z_1 = 1 + 2i \).

### Step-by-Step Solution:

1. **Identify the Complex Number**:
Let \( z = x + yi \), where \( x \) and \( y \) are real numbers. The given complex number is \( z_1 = 1 + 2i \).

2. **Calculate \( z - z_1 \)**:
\[ z - z_1 = (x + yi) - (1 + 2i) = (x - 1) + (y - 2)i \]

3. **Find the Modulus**:
The modulus \( |z - z_1| \) is given by:
\[ |z - z_1| = \sqrt{(x - 1)^2 + (y - 2)^2} \]

4. **Set Up the Inequality**:
We need to analyze the inequality:
\[ 1 < |z - z_1| \leq 3 \]
This translates to:
\[ 1 < \sqrt{(x - 1)^2 + (y - 2)^2} \leq 3 \]

5. **Square the Inequalities**:
Squaring both sides of the inequalities gives:
\[ 1^2 < (x - 1)^2 + (y - 2)^2 \leq 3^2 \]
Which simplifies to:
\[ 1 < (x - 1)^2 + (y - 2)^2 \leq 9 \]

6. **Interpret the Results**:
The inequality \( (x - 1)^2 + (y - 2)^2 \leq 9 \) represents a circle centered at \( (1, 2) \) with a radius of 3. The inequality \( 1 < (x - 1)^2 + (y - 2)^2 \) represents the area outside a circle centered at \( (1, 2) \) with a radius of 1.

7. **Determine the Region**:
Therefore, the region described by \( 1 < |z - z_1| \leq 3 \) is the area between the two circles:
- The inner circle (radius 1) is not included (open circle).
- The outer circle (radius 3) is included (closed circle).

### Argand Diagram Representation:
- Draw a coordinate system (the Argand diagram).
- Mark the center of the circles at the point \( (1, 2) \).
- Draw a circle with radius 3 centered at \( (1, 2) \) (this is included).
- Draw a circle with radius 1 centered at \( (1, 2) \) (this is not included).
- Shade the region between these two circles to represent the solution.