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If $i=\sqrt{-1}$, then $(1+i)^{10}+(1-i)^{10}=$
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$\begin{aligned} & (1+i)^{10}+(1-i)^{10} \\ = & 2^5\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^{10}+2^5\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)^{10} \\ = & 2^5\left[\left(e^{i \pi / 4}\right)^{10}+\left(e^{-i \pi / 4}\right)^{10}\right]\end{aligned}$
$\begin{aligned} & =2^5\left[e^{i 5 \pi / 2}+e^{-i 5 \pi / 2}\right] \\ & =2^5\left[e^{i \pi / 2}+e^{-i \pi / 2}\right] \\ & =2^5\left[i \sin \frac{\pi}{2}-i \sin \frac{\pi}{2}\right]=0 .\end{aligned}$
$\begin{aligned} & =2^5\left[e^{i 5 \pi / 2}+e^{-i 5 \pi / 2}\right] \\ & =2^5\left[e^{i \pi / 2}+e^{-i \pi / 2}\right] \\ & =2^5\left[i \sin \frac{\pi}{2}-i \sin \frac{\pi}{2}\right]=0 .\end{aligned}$
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