If `(x+iy)^(2)-7+24i`, then the value of `(7+sqrt(-576))^(1//2)-(7-sqrt(-576))^(1//2)`, is

Options

A. `-6i`
B. `-3i`
C. `2i`
D. 6

Explanation

To solve the problem, we need to evaluate the expression \( (7 + \sqrt{-576})^{1/2} - (7 - \sqrt{-576})^{1/2} \).

### Step-by-Step Solution:

1. **Identify the square roots of the complex numbers**:
\[ \sqrt{-576} = \sqrt{576} \cdot \sqrt{-1} = 24i \]
Therefore, we can rewrite our expression as:
\[ (7 + 24i)^{1/2} - (7 - 24i)^{1/2} \]

2. **Let \( z_1 = 7 + 24i \) and \( z_2 = 7 - 24i \)**:
We need to find \( \sqrt{z_1} \) and \( \sqrt{z_2} \).

3. **Calculate the modulus and argument of \( z_1 \)**:
- Modulus:
\[ |z_1| = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \]
- Argument:
\[ \theta_1 = \tan^{-1}\left(\frac{24}{7}\right) \]

4. **Express \( z_1 \) in polar form**:
\[ z_1 = 25 \left( \cos(\theta_1) + i \sin(\theta_1) \right) \]

5. **Find the square root of \( z_1 \)**:
\[ \sqrt{z_1} = \sqrt{25} \left( \cos\left(\frac{\theta_1}{2}\right) + i \sin\left(\frac{\theta_1}{2}\right) \right) = 5 \left( \cos\left(\frac{\theta_1}{2}\right) + i \sin\left(\frac{\theta_1}{2}\right) \right) \]

6. **Calculate the modulus and argument of \( z_2 \)**:
- Modulus:
\[ |z_2| = \sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \]
- Argument:
\[ \theta_2 = \tan^{-1}\left(\frac{-24}{7}\right) \]

7. **Express \( z_2 \) in polar form**:
\[ z_2 = 25 \left( \cos(\theta_2) + i \sin(\theta_2) \right) \]

8. **Find the square root of \( z_2 \)**:
\[ \sqrt{z_2} = 5 \left( \cos\left(\frac{\theta_2}{2}\right) + i \sin\left(\frac{\theta_2}{2}\right) \right) \]

9. **Now substitute back into the expression**:
\[ \sqrt{z_1} - \sqrt{z_2} = 5 \left( \cos\left(\frac{\theta_1}{2}\right) + i \sin\left(\frac{\theta_1}{2}\right) \right) - 5 \left( \cos\left(\frac{\theta_2}{2}\right) + i \sin\left(\frac{\theta_2}{2}\right) \right) \]

10. **Simplify the expression**:
The real parts cancel out, and we are left with:
\[ 5i \left( \sin\left(\frac{\theta_1}{2}\right) - \sin\left(\frac{\theta_2}{2}\right) \right) \]

11. **Final Result**:
Since the imaginary parts yield \( 6i \) after evaluation, we conclude:
\[ \text{The value is } 6i \]