We are asked to find the value of the expression:
\(\sqrt{-25} + 3\sqrt{-4} + 2\sqrt{-9}\)
We can simplify each square root term by recognizing that the square root of a negative number involves the imaginary unit \(i\), where \(i = \sqrt{-1} \)
1. \(\sqrt{-25} = \sqrt{25} \times \sqrt{-1} = 5i\)
2. \(3\sqrt{-4} = 3 \times \sqrt{4} \times \sqrt{-1} = 3 \times 2 \times i = 6i\)
3. \(2\sqrt{-9} = 2 \times \sqrt{9} \times \sqrt{-1} = 2 \times 3 \times i = 6i\)
Now, combine all the terms:
\(5i + 6i + 6i = 17i\)
The answer is \( 17i \).
If \( z \) is a complex number and \( k \in \mathbb{R} \), such that \( |z| = 1 \), \[ \frac{2 + k^2 z}{k + \overline{z}} = kz, \] then the maximum distance from \( k + i k^2 \) to the circle \( |z - (1 + 2i)| = 1 \) is: