if `|z-2i| le sqrt2` , then the maximum value of |3+i(z-1)| is :

Options

A. `sqrt2`
B. `2sqrt2`
C. `2+sqrt2`
D. `3+2sqrt2`

Answer

To solve the problem, we need to find the maximum value of \( |3 + i(z - 1)| \) given the condition \( |z - 2i| \leq \sqrt{2} \).

### Step-by-Step Solution:

1. **Understand the given condition**:
We have \( |z - 2i| \leq \sqrt{2} \). This represents a circle in the complex plane centered at \( 2i \) (which is the point \( (0, 2) \) in the Cartesian plane) with a radius of \( \sqrt{2} \).

2. **Express \( z \)**:
Let \( z = x + iy \). The condition can be rewritten as:
\[ |(x + iy) - 2i| \leq \sqrt{2} \implies |x + i(y - 2)| \leq \sqrt{2} \]
This implies:
\[ \sqrt{x^2 + (y - 2)^2} \leq \sqrt{2} \]
Squaring both sides gives:
\[ x^2 + (y - 2)^2 \leq 2 \]

3. **Rearranging the expression**:
We need to find the maximum value of \( |3 + i(z - 1)| \):
\[ |3 + i(z - 1)| = |3 + i(x + iy - 1)| = |3 + i(x - 1 + iy)| \]
This can be expressed as:
\[ |3 + i(x - 1 + iy)| = |3 + i((y - 2) + 2)| = |3 + i(y - 1)| \]

4. **Using the triangle inequality**:
We can use the triangle inequality:
\[ |3 + i(y - 1)| \leq |3| + |i(y - 1)| = 3 + |y - 1| \]

5. **Finding the maximum of \( |y - 1| \)**:
From the condition \( x^2 + (y - 2)^2 \leq 2 \), we can find the maximum and minimum values of \( y \):
- The center of the circle is at \( (0, 2) \) with a radius of \( \sqrt{2} \).
- The maximum value of \( y \) occurs at \( y = 2 + \sqrt{2} \) and the minimum value occurs at \( y = 2 - \sqrt{2} \).

6. **Calculating \( |y - 1| \)**:
- Maximum \( y = 2 + \sqrt{2} \):
\[ |y - 1| = |(2 + \sqrt{2}) - 1| = |1 + \sqrt{2}| \]
- Minimum \( y = 2 - \sqrt{2} \):
\[ |y - 1| = |(2 - \sqrt{2}) - 1| = |1 - \sqrt{2}| \]
- The maximum value of \( |y - 1| \) occurs at \( y = 2 + \sqrt{2} \):
\[ |y - 1| = 1 + \sqrt{2} \]

7. **Final calculation**:
Now substituting back into our inequality:
\[ |3 + i(y - 1)| \leq 3 + (1 + \sqrt{2}) = 4 + \sqrt{2} \]

### Conclusion:
Thus, the maximum value of \( |3 + i(z - 1)| \) is \( 4 + \sqrt{2} \).