This one seems pretty simple, we are asked to represent $(\sqrt{3}+i)^{100}$ as a complex number of the form $a+bi$ but maybe I'm underthinking it..Taking advantage of binomial properties, I get:
$$(\sqrt{3}+i)^{100}=(3^{50}-1) + (100\sqrt{3})i=\left|(3^{50}-1) + (100\sqrt{3})i\right| e^{tan^{-1}(\frac{(100\sqrt{3})}{(3^{50}-1)})i}$$
So $a=3^{50}-1$ and $b=100\sqrt{3}$.
Could you please state the problem so I can know if I like this or not? I'm on my iPhone at a restaurant with a dog and my kid and stuff so clicking over to check seems like more of a pain and less fun than complaining. 
Also, I was thinking on terms of $re^{i\theta}$ rather than the binomial theorem, though I think that could work as well.
This will probably simplify and there is probably some kind of shortcut I missed but I think it is still accurate.
I suppose I could just go with:
$$(\sqrt{3}+i)^{100}=(2e^{\frac{\pi}{6}})^{100}=\left(2^{100}\right)e^{\frac{100\pi}{6}}= \left(2^{100}\right)e^{\frac{2\pi}{3}}$$
Then I can decompose this into rectangular coordinates:
$$-2^{99} + \left( ( 2^{99}) \sqrt{3} \right) i$$
@hjoseph As I said, please state the problem. If you do, you will see the problem with that last answer. Also (and I'm really mentioning this more so that you can learn stack-type etiquette) I think it's generally a good idea to edit and improve your original post, rather than add more stuff.
This seemed better in my head but upon second thought I don't think binomials have the property I used...
I think to rewrite in a+bi form you can just find the polar form of the complex polynomial and use that theta in a+bi.
$$\begin{align}
\quad(\sqrt{3}+i)^{100}\\\
\quad r=2, \tan^{-1}\frac{1}{\sqrt{3}}=\frac{\pi}{6} \quad \\\
(2)^{100}e^\frac{100 \pi i}{6} \\\
2^{100}\cos(\frac{100 \pi}{6})+2^{100} \sin(\frac{100 \pi}{6})i \quad
\end{align}$$
The final answer is equal to the original value and is in the proper form.
I did make the mistake of putting the reciprocal of the $tangent$ in my $arctan$, after some minor adjustments, I think what I have is equivalent.
I think I like Matt's the best so far - though, I think it's not too hard to recognize the sine and cosine of $100\pi/6$.