$\frac{2(1+i)(1-i\sqrt{3})}{(-\sqrt{3}+i)^{30}}$
So I solved this by using polar form and then turned it to cis
for (1+i) the argument is $\sqrt{2}$exp($\frac{\pi}{4}$i)
for (1-i$\sqrt{3}$) the argument is 2exp($\frac{-\pi}{3}$i)
for $(-\sqrt{3}+i)^{30}$ the argument is 2^${30}$exp(25$\pi$i)
then total argument of this 2^(${\frac{-55}{2}}$)exp($\frac{-301\pi}{12}$) then I turned this into cis which will be 2^(${\frac{-55}{2}}$)$\cos(\frac{-301\pi}{12})$+i$\sin(\frac{-301\pi}{12})$ then argument is from boundary of -$\pi$ < $\theta$ $\leq$ $\pi$
so ($\frac{-301\pi}{12})$ + k$\pi$ (k = 26) = $\frac{11\pi}{12}$
answer is 2^(${\frac{-55}{2}}$)$\cos(\frac{11\pi}{12})$+i$\sin(\frac{11\pi}{12})$
the book has it as this format:
$\frac{2^({\frac{-55}{2}})\cos(\frac{-\pi}{12})+i\sin(\frac{-\pi}{12})}{\cos({25\pi})+i\sin({25\pi)}}$
and that equals to -2^(${\frac{-55}{2}}$)$\cos(\frac{-\pi}{12})$+i$\sin(\frac{-\pi}{12})$ and then somehow they got the same answer as me, what happened to the negative in the front and how it turned into positive. I hope my work is correctly done via polar form.