Find the imaginary part of `((3+2sqrt(-54))^(1/2)-(3-2sqrt(-54))^(1/2))`

Options

A. `-sqrt(6)`
B. `-2sqrt(6)`
C. `sqrt(6)`
D. 6

Answer

To find the imaginary part of the expression \( \left( (3 + 2\sqrt{-54})^{1/2} - (3 - 2\sqrt{-54})^{1/2} \right) \), we will follow these steps:

### Step 1: Simplify the square roots
First, we rewrite the square roots of the complex numbers. Since \( \sqrt{-54} = \sqrt{54}i = 3\sqrt{6}i \), we can rewrite the expression as:
\[ 3 + 2\sqrt{-54} = 3 + 2(3\sqrt{6}i) = 3 + 6\sqrt{6}i \]
\[ 3 - 2\sqrt{-54} = 3 - 2(3\sqrt{6}i) = 3 - 6\sqrt{6}i \]

### Step 2: Define the complex numbers
Let:
\[ z_1 = 3 + 6\sqrt{6}i \]
\[ z_2 = 3 - 6\sqrt{6}i \]

### Step 3: Find the square roots of \( z_1 \) and \( z_2 \)
To find the square roots of a complex number \( z = a + bi \), we use the formula:
\[ \sqrt{z} = \sqrt{r} \left( \cos\left(\frac{\theta}{2}\right) + i\sin\left(\frac{\theta}{2}\right) \right) \]
where \( r = \sqrt{a^2 + b^2} \) and \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).

Calculating \( r_1 \) for \( z_1 \):
\[ r_1 = \sqrt{3^2 + (6\sqrt{6})^2} = \sqrt{9 + 216} = \sqrt{225} = 15 \]

Calculating \( r_2 \) for \( z_2 \):
\[ r_2 = \sqrt{3^2 + (-6\sqrt{6})^2} = \sqrt{9 + 216} = \sqrt{225} = 15 \]

### Step 4: Find the angles
For \( z_1 \):
\[ \theta_1 = \tan^{-1}\left(\frac{6\sqrt{6}}{3}\right) = \tan^{-1}(2\sqrt{6}) \]

For \( z_2 \):
\[ \theta_2 = \tan^{-1}\left(\frac{-6\sqrt{6}}{3}\right) = \tan^{-1}(-2\sqrt{6}) \]

### Step 5: Calculate the square roots
Thus, we have:
\[ \sqrt{z_1} = \sqrt{15} \left( \cos\left(\frac{\theta_1}{2}\right) + i\sin\left(\frac{\theta_1}{2}\right) \right) \]
\[ \sqrt{z_2} = \sqrt{15} \left( \cos\left(\frac{\theta_2}{2}\right) + i\sin\left(\frac{\theta_2}{2}\right) \right) \]

### Step 6: Subtract the square roots
Now we need to find:
\[ \sqrt{z_1} - \sqrt{z_2} \]
This will give us a complex number whose imaginary part we need to find.

### Step 7: Find the imaginary part
The imaginary part of \( \sqrt{z_1} - \sqrt{z_2} \) can be calculated as:
\[ \text{Imaginary part} = \sqrt{15}\left(\sin\left(\frac{\theta_1}{2}\right) - \sin\left(\frac{\theta_2}{2}\right)\right) \]

Since \( \sin\left(\frac{\theta_2}{2}\right) = -\sin\left(\frac{\theta_1}{2}\right) \), we have:
\[ \text{Imaginary part} = \sqrt{15}\left(\sin\left(\frac{\theta_1}{2}\right) + \sin\left(\frac{\theta_1}{2}\right)\right) = 2\sqrt{15}\sin\left(\frac{\theta_1}{2}\right) \]

### Conclusion
Thus, the imaginary part of the expression \( \left( (3 + 2\sqrt{-54})^{1/2} - (3 - 2\sqrt{-54})^{1/2} \right) \) is \( 2\sqrt{15}\sin\left(\frac{\theta_1}{2}\right) \).