Question

已知i为虚数单位,求(2+2i)^4/(1-√3i)^5的虚部

Answer 1

(2+2i)^4/(1-√3i)^5
=[(2^4)(1+i)^4]/[(-2)(-1/2+√3/2i)]^5
=[(2^4)(2i)^2]/[(-2^5)(-1/2+√3/2i)^2]
=-64/[(-32)(-1/2-√3/2i)]
=2/(-1/2-√3/2i)
=-4/(1+√3i)
=-4(1-√3i)/4
=-1+√3i
所以虚部是-1