[No.01]

1+\(\sqrt{\quad}\)3・i=2(cos60°+i・sin60°)
(1+\(\sqrt{\quad}\)3・i\()^{10}\)=(\(2^{10}\))(cos600°+i・sin600°）【ド・モアブルの定理】
=(\(2^{10}\))(-\(\frac{1}{2}\)-\(\sqrt{\quad}\)\(\frac{3}{2}\)・i)
=-(\(2^{9}\))(1+\(\sqrt{\quad}\)3・i)
=-512(1+\(\sqrt{\quad}\)3・i)
その逆数だから
1/-512(1+\(\sqrt{\quad}\)3・i)
=(\(\sqrt{\quad}\)3・i-1)/2048

[No.02]

8^(\(\frac{1}{6}\))=(\(2^{3}\))^(\(\frac{1}{6}\))=2^(\(\frac{1}{2}\))=\(\sqrt{\quad}\)2

-4\(\sqrt{\quad}\)2+4\(\sqrt{\quad}\)2・i=8{(-\(\sqrt{\quad}\)\(\frac{2}{2}\))+(\(\sqrt{\quad}\)\(\frac{2}{2}\))・i}
=(\(2^{3}\))(cos135°+i・sin135°)
=(\(2^{3}\)){cos(3×45°)+i・sin(3×45°)}
=(\(2^{3}\))(cos45°+i・sin45°\()^{3}\)
よって
(-4\(\sqrt{\quad}\)2+4\(\sqrt{\quad}\)2・i)^(\(\frac{1}{3}\)）
=2(cos45°+i・sin45°)
=\(\sqrt{\quad}\)2(1+i)