（3-4i/4+3i）+（1-i/1+i）的199次方
问题描述：
（3-4i/4+3i）+（1-i/1+i）的199次方
答
(3-4i)/(4+3i)
=(3-4i)(4-3i)/(4+3i)(4-3i)
=(12-9i-16i+12i²)/(16+9)
=25i/25
=i
(1-i)/(1+i)
=(1-i)²/(1+i)(1-i)
=(1-2i+i²)/(1+1)
=-2i/2
=-i
原式=i+(-1)^199
=i-i^199
=i-(-i)
=2i