`((-1+sqrt(3)i)/(1-i))^(30)` simplifies to

Options

A. `-2^(15)i`
B. `2^(15)i`
C. `2^(15)`
D. `-2^(15)`

Answer

To simplify the expression \(\left(\frac{-1+\sqrt{3}i}{1-i}\right)^{30}\), we will convert the complex numbers into polar form and then apply De Moivre's theorem.

### Step 1: Convert the numerator \(-1 + \sqrt{3}i\) to polar form
1. **Calculate the modulus**:
\[ r = |z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

2. **Calculate the argument**:
\[ \theta = \tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right) \]
Since the point \((-1, \sqrt{3})\) is in the second quadrant, we can find the angle:
\[ \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

3. **Express in polar form**:
\[ -1 + \sqrt{3}i = 2 \left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 2e^{i\frac{2\pi}{3}} \]

### Step 2: Convert the denominator \(1 - i\) to polar form
1. **Calculate the modulus**:
\[ r = |z| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

2. **Calculate the argument**:
\[ \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \]

3. **Express in polar form**:
\[ 1 - i = \sqrt{2} \left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right) = \sqrt{2} e^{-i\frac{\pi}{4}} \]

### Step 3: Combine the fractions
Now we can write the original expression:
\[ \frac{-1+\sqrt{3}i}{1-i} = \frac{2e^{i\frac{2\pi}{3}}}{\sqrt{2}e^{-i\frac{\pi}{4}}} \]
This simplifies to:
\[ \frac{2}{\sqrt{2}} e^{i\left(\frac{2\pi}{3} + \frac{\pi}{4}\right)} = \sqrt{2} e^{i\left(\frac{2\pi}{3} + \frac{\pi}{4}\right)} \]

### Step 4: Find a common denominator for the angles
To add the angles:
\[ \frac{2\pi}{3} = \frac{8\pi}{12}, \quad \frac{\pi}{4} = \frac{3\pi}{12} \]
Thus,
\[ \frac{2\pi}{3} + \frac{\pi}{4} = \frac{8\pi}{12} + \frac{3\pi}{12} = \frac{11\pi}{12} \]
So we have:
\[ \frac{-1+\sqrt{3}i}{1-i} = \sqrt{2} e^{i\frac{11\pi}{12}} \]

### Step 5: Raise to the power of 30
Now we raise the expression to the power of 30:
\[ \left(\sqrt{2} e^{i\frac{11\pi}{12}}\right)^{30} = (\sqrt{2})^{30} \cdot e^{i\frac{30 \cdot 11\pi}{12}} = 2^{15} \cdot e^{i\frac{330\pi}{12}} = 2^{15} \cdot e^{i\frac{55\pi}{2}} \]

### Step 6: Simplify the angle
Now, simplify \(\frac{55\pi}{2}\):
\[ \frac{55\pi}{2} = 27\pi + \frac{\pi}{2} = 27\pi + \frac{\pi}{2} \equiv \frac{\pi}{2} \text{ (mod } 2\pi\text{)} \]

### Step 7: Final expression
Thus, we have:
\[ 2^{15} e^{i\frac{\pi}{2}} = 2^{15} \left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 2^{15} \cdot (0 + i) = 2^{15} i \]

### Conclusion
The final result is:
\[ \left(\frac{-1+\sqrt{3}i}{1-i}\right)^{30} = 2^{15} i \]