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 \( 17i \).
Let \( z \) satisfy \( |z| = 1, \ z = 1 - \overline{z} \text{ and } \operatorname{Im}(z)>0 \)
Then consider:
Statement-I: \( z \) is a real number
Statement-II: Principal argument of \( z \) is \( \dfrac{\pi}{3} \)
Then:
If \( z \) and \( \omega \) are two non-zero complex numbers such that \( |z\omega| = 1 \) and
\[ \arg(z) - \arg(\omega) = \frac{\pi}{2}, \]
Then the value of \( \overline{z\omega} \) is: