For a complex number Z, if `|Z-i|le2 and Z_(1)=5+3i`, then the maximum value of `|iZ+Z_(1)|` is (where, `i^(2)=-1`)

Options

A. `5+sqrt13`
B. `5+sqrt2`
C. 7
D. 8

Answer

To solve the problem, we need to find the maximum value of \( |iZ + Z_1| \) given the constraints on the complex number \( Z \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We have \( |Z - i| \leq 2 \) and \( Z_1 = 5 + 3i \). This means that the complex number \( Z \) lies within or on the boundary of a circle in the complex plane centered at \( i \) (which is \( 0 + 1i \)) with a radius of 2.

2. **Expressing \( |iZ + Z_1| \)**:
We want to find \( |iZ + Z_1| \). We can rewrite this as:
\[ |iZ + (5 + 3i)| = |iZ + 5 + 3i| = |iZ + 3i + 5| = |i(Z + 3) + 5| \]

3. **Finding the Modulus**:
We can express this as:
\[ |i(Z + 3) + 5| = |i(Z + 3)| + |5| \]
Using the triangle inequality, we have:
\[ |i(Z + 3)| = |Z + 3| \]

4. **Finding the Maximum of \( |Z + 3| \)**:
To find \( |Z + 3| \), we need to express \( Z \) in terms of its real and imaginary parts. Let \( Z = x + yi \). Then:
\[ |Z - i| = |(x + (y - 1)i)| \leq 2 \]
This implies:
\[ \sqrt{x^2 + (y - 1)^2} \leq 2 \]
Therefore, we have:
\[ x^2 + (y - 1)^2 \leq 4 \]
This describes a circle centered at \( (0, 1) \) with a radius of 2.

5. **Transforming the Circle**:
We want to maximize \( |Z + 3| = |(x + 3) + yi| = \sqrt{(x + 3)^2 + y^2} \). The center of this new expression is at \( (-3, 0) \).

6. **Finding the Distance**:
The distance from the center of the circle \( (0, 1) \) to the point \( (-3, 0) \) is:
\[ d = \sqrt{(-3 - 0)^2 + (0 - 1)^2} = \sqrt{9 + 1} = \sqrt{10} \]
The maximum value of \( |Z + 3| \) occurs at the farthest point on the circle from \( (-3, 0) \), which is:
\[ \text{Maximum } |Z + 3| = d + \text{radius} = \sqrt{10} + 2 \]

7. **Final Calculation**:
Therefore, the maximum value of \( |iZ + Z_1| \) is:
\[ |iZ + Z_1| = |Z + 3| + |5| = (\sqrt{10} + 2) + 5 = \sqrt{10} + 7 \]

### Conclusion:
Thus, the maximum value of \( |iZ + Z_1| \) is \( 7 + \sqrt{10} \).