(1+√3i)²=1+2√3i+3i²=-2+2√3i
(1+√3i)⁴={(1+√3i)²}²=(-2+2√3i)²
=4-8√3i+12i²
=-8-8√3i
(1+√3i)⁵=(1+√3i)⁴•(1+√3i)
=(-8-8√3i)(1+√3i)
=-8-8√3i-8√3i-24i²
=16-16√3i
{(1+√3i)/4}⁵=(16-16√3i)/4⁵
=4²(1-√3i)/4⁵
=(1-√3i)/4³
{(1+√3i)/4}⁻⁵=1/{(1+√3i)/4}⁵
=1/{(1-√3i)/4³}
=4³/(1-√3i)
分母分子に(1+√3i)かけて 有理化して
=4³(1+√3i)/(1-3i²)
=4³(1+√3i)/4
=4²(1+√3i)
=16+16√3i