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Do the following:
(a) Use exponential form to compute
i. $(1+\sqrt{3} i)^{2011}$;
$$\text { ii. }(1+\sqrt{3} i)^{-2011} \text { . }$$
(b) Prove that
$$\sum_{m=0}^{1005}\left(\begin{array}{c} 2011 \\ 2 m \end{array}\right)(-3)^{m}=2^{2010}$$
and
$$\sum_{m=0}^{1005}\left(\begin{array}{c} 2011 \\ 2 m+1 \end{array}\right)(-3)^{m}=2^{2010}$$
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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
$$\sum_{m=0}^{1005}\left(\begin{array}{c} 2011 \\ 2 m \end{array}\right)(-3)^{m}=\sum_{m=0}^{1005}\left(\begin{array}{c} 2011 \\ 2 m+1 \end{array}\right)(-3)^{m}=2^{2010}$$
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