The value of $$\left(\frac{-1+i\sqrt{3}}{1-i}\right)^{30}$$ is:

We have to evaluate the expression $$\left(\dfrac{-1+i\sqrt{3}}{1-i}\right)^{30}.$$ The most systematic way is to convert both the numerator and the denominator into polar (trigonometric) form, simplify the quotient, and finally apply De Moivre’s theorem for the 30-th power.

For any complex number $$z=a+ib$$ the modulus and the argument are given by the relations

$$r=\sqrt{a^{2}+b^{2}}, \qquad \theta=\tan^{-1}\!\left(\dfrac{b}{a}\right),$$

so that $$z=r\bigl(\cos\theta+i\sin\theta\bigr).$$

First we treat the numerator $$-1+i\sqrt{3}\,.$$

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

Because the real part is negative and the imaginary part is positive, the point lies in the second quadrant. There $$\tan\theta=\dfrac{\sqrt{3}}{-1}=-\sqrt{3}\;,$$ hence $$\theta=120^{\circ}= \dfrac{2\pi}{3}.$$

Thus

$$-1+i\sqrt{3}=2\bigl(\cos\tfrac{2\pi}{3}+i\sin\tfrac{2\pi}{3}\bigr).$$

Next we examine the denominator $$1-i.$$

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

Here the real part is positive and the imaginary part is negative, placing the point in the fourth quadrant. We have $$\tan\phi=\dfrac{-1}{1}=-1,$$ so $$\phi=-45^{\circ}= -\dfrac{\pi}{4}.$$ Therefore

$$1-i=\sqrt{2}\bigl(\cos(-\tfrac{\pi}{4})+i\sin(-\tfrac{\pi}{4})\bigr).$$

Now we divide the two complex numbers. Using the fact that

$$\dfrac{r_{1}(\cos\theta_{1}+i\sin\theta_{1})}{r_{2}(\cos\theta_{2}+i\sin\theta_{2})} =\dfrac{r_{1}}{r_{2}}\bigl(\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})\bigr),$$

we find

$$\dfrac{-1+i\sqrt{3}}{1-i} =\dfrac{2}{\sqrt{2}}\Bigl(\cos\!\Bigl(\tfrac{2\pi}{3}-(-\tfrac{\pi}{4})\Bigr) +i\sin\!\Bigl(\tfrac{2\pi}{3}-(-\tfrac{\pi}{4})\Bigr)\Bigr).$$

The radial part is $$\dfrac{2}{\sqrt{2}}=\sqrt{2}.$$

The angular part is

$$\tfrac{2\pi}{3}-(-\tfrac{\pi}{4})=\tfrac{2\pi}{3}+\tfrac{\pi}{4} =\dfrac{8\pi}{12}+\dfrac{3\pi}{12} =\dfrac{11\pi}{12}.$$

Hence the quotient is

$$\sqrt{2}\bigl(\cos\tfrac{11\pi}{12}+i\sin\tfrac{11\pi}{12}\bigr).$$

We must now raise this to the 30-th power. By De Moivre’s theorem,

$$(r(\cos\theta+i\sin\theta))^{n}=r^{n}\bigl(\cos n\theta+i\sin n\theta\bigr).$$

Taking $$r=\sqrt{2}$$, $$\theta=\dfrac{11\pi}{12}$$ and $$n=30$$, we obtain

$$\Bigl(\sqrt{2}(\cos\tfrac{11\pi}{12}+i\sin\tfrac{11\pi}{12})\Bigr)^{30} =(\sqrt{2})^{30}\Bigl(\cos(30\cdot\tfrac{11\pi}{12})+i\sin(30\cdot\tfrac{11\pi}{12})\Bigr).$$

First compute the modulus:

$$(\sqrt{2})^{30}=(2^{1/2})^{30}=2^{15}.$$

Next compute the angle:

$$30\cdot\dfrac{11\pi}{12}=\dfrac{330\pi}{12}=\dfrac{55\pi}{2}.$$

Because a full revolution is $$2\pi,$$ we subtract as many $$2\pi$$ as possible:

$$\dfrac{55\pi}{2}-13\cdot 2\pi =\dfrac{55\pi}{2}-\dfrac{52\pi}{2} =\dfrac{3\pi}{2}.$$

Therefore the required power equals

$$2^{15}\bigl(\cos\tfrac{3\pi}{2}+i\sin\tfrac{3\pi}{2}\bigr).$$

We know the exact trigonometric values

$$\cos\tfrac{3\pi}{2}=0,\qquad \sin\tfrac{3\pi}{2}=-1.$$

Substituting, we get

$$2^{15}(0-i)= -\,2^{15}i.$$

Hence, the correct answer is Option D.