已知:z=[(2-i)(4i-2)]/1+i,求复数z的共轭复数的三角形式

Question

已知:z=[(2-i)(4i-2)]/1+i,求复数z的共轭复数的三角形式

Answer 1

z=[(2-i)(4i-2)]/1+i
=10i/(1+i)
=5+5i
所以z的共轭复数
=5-5i
=5√2[cos(7π/4)+isin(7π/4)]