Solution. Since

$$

1+\sqrt{3} i=2\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=2 \exp \left(\frac{\pi i}{3}\right)

$$

we have

$$

\begin{aligned}

(1+\sqrt{3} i)^{2011} &=2^{2011} \exp \left(\frac{2011 \pi i}{3}\right)=2^{2011} \exp \left(\frac{2011 \pi i}{3}\right) \\

&=2^{2011} \exp \left(670 \pi i+\frac{\pi i}{3}\right) \\

&=2^{2011} \exp \left(\frac{\pi i}{3}\right)=2^{2011}\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right) \\

&=2^{2010}(1+\sqrt{3} i) .

\end{aligned}

$$

Similarly,

$$

(1+\sqrt{3} i)^{-2011}=2^{-2013}(1-\sqrt{3} i)

$$

$$

\begin{aligned}

2^{2010}(1+\sqrt{3} i) &=(1+\sqrt{3} i)^{2011} \\

&=\sum_{m=0}^{1005}\left(\begin{array}{c}

2011 \\

2 m

\end{array}\right)(-3)^{m}+i \sum_{m=0}^{1005}\left(\begin{array}{c}

2011 \\

2 m+1

\end{array}\right)(-3)^{m} \sqrt{3} .

\end{aligned}

$$

It follows that

$$

\sum_{m=0}^{1005}\left(\begin{array}{c}

2011 \\

2 m

\end{array}\right)(-3)^{m}=\sum_{m=0}^{1005}\left(\begin{array}{c}

2011 \\

2 m+1

\end{array}\right)(-3)^{m}=2^{2010}

$$