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. **Understanding the Condition**:
The condition \( |z - 2i| \leq \sqrt{2} \) describes a circle in the complex plane. The center of this circle is at \( 0 + 2i \) (which corresponds to the point \( (0, 2) \) in the Cartesian plane) and has a radius of \( \sqrt{2} \).

2. **Equation of the Circle**:
The equation of the circle can be expressed as:
\[ |z - 2i|^2 \leq 2 \]
If we let \( z = x + yi \), then:
\[ |(x + yi) - 2i|^2 = |x + (y - 2)i|^2 = x^2 + (y - 2)^2 \leq 2 \]
This simplifies to:
\[ x^2 + (y - 2)^2 \leq 2 \]

3. **Finding the Expression**:
We need to analyze the expression \( |3 + i(z - 1)| \):
\[ |3 + i(z - 1)| = |3 + i(x + yi - 1)| = |3 + i(x - 1 + yi)| \]
This can be rewritten as:
\[ |3 + i(y - 1 + x)| = |3 + i(y - 1)| \]

4. **Distance from the Point**:
We can rewrite \( |3 + i(z - 1)| \) as:
\[ |3 + i(z - 1)| = |3 + i(y - 1 + x)| \]
This represents the distance from the point \( (3, 0) \) to the point \( (x, y) \) where \( z = x + yi \).

5. **Finding Maximum Distance**:
The maximum distance from the point \( (3, 0) \) to any point on the circle centered at \( (0, 2) \) with radius \( \sqrt{2} \) occurs when the two points are diametrically opposite each other. The distance from the center of the circle \( (0, 2) \) to \( (3, 0) \) can be calculated as:
\[ \text{Distance} = \sqrt{(3 - 0)^2 + (0 - 2)^2} = \sqrt{9 + 4} = \sqrt{13} \]
Therefore, the maximum distance from \( (3, 0) \) to any point on the circle is:
\[ \text{Maximum Distance} = \sqrt{13} + \sqrt{2} \]

6. **Final Calculation**:
The maximum value of \( |3 + i(z - 1)| \) is therefore:
\[ \text{Maximum Value} = \sqrt{13} + \sqrt{2} \]

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