1. We need to compute

$$\left(\frac{-1+i\sqrt{3}}{1-i}\right)^{30}.$$

2. First, write numerator and denominator in polar form.

Numerator: $-1+i\sqrt{3}$

Its modulus is

$$\sqrt{(-1)^2+(\sqrt{3})^2} = \sqrt{1+3}=2.$$

Its argument lies in the second quadrant, and

$$\tan \theta = \frac{\sqrt{3}}{-1}=-\sqrt{3}.$$

So the principal argument is

$$\theta = \frac{2\pi}{3}.$$

Hence,

$$-1+i\sqrt{3} = 2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right).$$

Denominator: $1-i$

Its modulus is

$$\sqrt{1^2+(-1)^2}=\sqrt{2}.$$

Its argument is

$$-\frac{\pi}{4}.$$

Thus,

$$1-i = \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right).$$

3. Now divide the two complex numbers.

The modulus becomes

$$\frac{2}{\sqrt{2}}=\sqrt{2}.$$

The argument becomes

\frac{2\pi}{3}-\left(-\frac{\pi}{4}\right)=\frac{2\pi}{3}+\frac{\pi}{4}=rac{8\pi+3\pi}{12}=rac{11\pi}{12}.

So,

$$\frac{-1+i\sqrt{3}}{1-i}=\sqrt{2}\left(\cos\frac{11\pi}{12}+i\sin\frac{11\pi}{12}\right).$$

4. Raise to the 30th power using De Moivre's theorem:

$$\left(\sqrt{2}\right)^{30}\left(\cos\frac{11\pi}{12}+i\sin\frac{11\pi}{12}\right)^{30} =2^{15}\left(\cos\frac{55\pi}{2}+i\sin\frac{55\pi}{2}\right).$$

Since

$$2^{15}=32768,$$

and

$$\frac{55\pi}{2}=27\pi+\frac{\pi}{2},$$

we get

$$\cos\left(27\pi+\frac{\pi}{2}\right)=0, \qquad \sin\left(27\pi+\frac{\pi}{2}\right)=-1.$$

Therefore,

$$2^{15}\left(0-i\right)=-32768i.$$

5. So the value is

$$\boxed{-32768i}.$$

6. Compare with options:

• A: $-215i$
• B: $-215$
• C: $215i$
• D: $65$

None of the options match $-32768i$. So the stored correct answer A is incorrect.