Answer
$\color{blue}{{\bf d.}\ \frac{1}{4} - i\frac{\sqrt{3}}{4}} $
Work Step by Step
Let $z_1= 1 - i,\ z_2 = -\sqrt{3} - i,\ z_3 = 1 +i\sqrt{3}$.
We wish to compute
$\dfrac{z_1^4z_2^2}{z_3^5} = \dfrac{(1-i)^4(-\sqrt{3}-i)^2}{(1+i\sqrt{3})^5}$.
Converting $z_1, z_2$, and $z_3$ to trigonometric form:
$\scriptsize\begin{array}{|l|c|c|c|c|c|} \hline
\hfill z\hfill & x & y & r & \tan\theta,\ \theta & \text{Trig. Form} \\ \hline
z_1 = 1 - i & 1 & -1 & \sqrt{1^2+(-1)^2} & \tan\theta = -1/1 = -1 & \\
&&& = \sqrt{2} & \theta\in QIV, \theta= 7\pi/4 & z_1=\sqrt{2}\ \text{cis}\ 7\pi/4 \\ \hline
z_2 = -\sqrt{3} - i & -\sqrt{3} & -1 & \sqrt{(-\sqrt{3})^2+(-1)^2} & \tan\theta = -1/(-\sqrt{3}) = \sqrt{3}/3 & \\
&&& = \sqrt{4} = 2 & \theta\in QIII, \theta= 7\pi/6 & z_2=2\ \text{cis}\ 7\pi/6 \\ \hline
z_3 = 1 +i\sqrt{3} & 1 & \sqrt{3} & \sqrt{1^2 + (\sqrt{3})^2} & \tan\theta = \sqrt{3}/1 = \sqrt{3} & \\
&&& = \sqrt{4} = 2 & \theta\in QI, \theta= \pi/3 & z_2=2\ \text{cis}\ \pi/3 \\ \hline
\end{array}$
Compute:
$\begin{align*}
\dfrac{z_1^4z_2^2}{z_3^5} &= \dfrac{(1-i)^4(-\sqrt{3}-i)^2}{(1+i\sqrt{3})^5} \\
&= \dfrac{(\sqrt{2}\ \text{cis}\ 7\pi/4)^4(2\ \text{cis}\ 7\pi/6)^2}{(2\ \text{cis}\ \pi/3)^5}
&\text{[see table]}\\
&= \dfrac{\left((\sqrt{2})^4\ \text{cis}\ 4(7\pi/4)\right) \left(2^2\ \text{cis}\ 2(7\pi/6)\right)}{2^5\ \text{cis}\ 5(\pi/3)} & \text{[De Moivre's Theorem]}\\
&= \dfrac{(4\ \text{cis}\ 7\pi)(4\ \text{cis}\ 7\pi/3)}{32\ \text{cis}\ 5\pi/3} \\
&= \dfrac{\left(4\ \text{cis}\ (6\pi+\pi)\right)\left(4\ \text{cis}\ (2\pi+\pi/3)\right)}{32\ \text{cis}\ 5\pi/3} \\
&= \dfrac{(4\ \text{cis}\ \pi)(4\ \text{cis}\ \pi/3)}{32\ \text{cis}\ 5\pi/3} \\
&= \dfrac{(4)(4)}{32}\ \text{cis}\ (\pi + \pi/3 -5\pi/3) & \text{[Mult./Div. Trig. Form]}\\
&= \dfrac{1}{2}\ \text{cis}\ (-\pi/3) \\
&= \dfrac{1}{2}(\cos(-\pi/3) + i\sin(-\pi/3)) \\
&= \dfrac{1}{2}(1/2 - i\sqrt{3}/2) \\
\color{blue}{\dfrac{(1-i)^4(-\sqrt{3}-i)^2}{(1+i\sqrt{3})^5}} &\color{blue}{= \dfrac{1}{4} - i\dfrac{\sqrt{3}}{4} \qquad \text{(Choice}\ {\bf d.})}
\end{align*}$
