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 step by step, we will start from the given equation and work through the calculations systematically.

### Step 1: Start with the given equation
We are given that:
\[ (x + iy)^2 = 7 + 24i \]

### Step 2: Take the square root of both sides
Taking the square root of both sides, we have:
\[ x + iy = \sqrt{7 + 24i} \]

### Step 3: Find the expression we need to evaluate
We need to evaluate the expression:
\[ (7 + \sqrt{-576})^{1/2} - (7 - \sqrt{-576})^{1/2} \]

### Step 4: Simplify \(\sqrt{-576}\)
We know that \(\sqrt{-576} = \sqrt{576} \cdot \sqrt{-1} = 24i\). Thus, we can rewrite the expression as:
\[ (7 + 24i)^{1/2} - (7 - 24i)^{1/2} \]

### Step 5: Use the square root of a complex number
Let \(z_1 = 7 + 24i\) and \(z_2 = 7 - 24i\). We can find the square roots of these complex numbers.

### Step 6: Calculate the modulus and argument of \(z_1\)
The modulus of \(z_1\) is:
\[ |z_1| = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \]
The argument \(\theta\) can be calculated using:
\[ \tan \theta = \frac{24}{7} \]

### Step 7: Find the square root of \(z_1\)
The square root of a complex number in polar form is given by:
\[ \sqrt{r} \left( \cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) \right) \]
Thus:
\[ \sqrt{7 + 24i} = \sqrt{25} \left( \cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) \right) = 5 \left( \cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) \right) \]

### Step 8: Find the square root of \(z_2\)
Similarly, for \(z_2\):
\[ \sqrt{7 - 24i} = \sqrt{25} \left( \cos\left(-\frac{\theta}{2}\right) + i \sin\left(-\frac{\theta}{2}\right) \right) = 5 \left( \cos\left(-\frac{\theta}{2}\right) + i \sin\left(-\frac{\theta}{2}\right) \right) \]

### Step 9: Subtract the two square roots
Now we can compute:
\[ \sqrt{7 + 24i} - \sqrt{7 - 24i} = 5 \left( \cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) \right) - 5 \left( \cos\left(-\frac{\theta}{2}\right) + i \sin\left(-\frac{\theta}{2}\right) \right) \]
This simplifies to:
\[ 5 \left( 0 + i \left( \sin\left(\frac{\theta}{2}\right) - (-\sin\left(\frac{\theta}{2}\right)) \right) \right) = 10i \sin\left(\frac{\theta}{2}\right) \]

### Step 10: Find the value of \(y\)
From the original equation, we had:
\[ 2iy = 10i \sin\left(\frac{\theta}{2}\right) \]
Thus:
\[ y = 5 \sin\left(\frac{\theta}{2}\right) \]

### Step 11: Final value
Since we are looking for the imaginary part, we conclude:
\[ \text{The value is } 6i \text{ or } -6i \]

### Conclusion
The final answer is:
\[ -6i \]