来自程志华的问题
虚数化简i^(7x+2)还有i^(6x-1),i^(14x-11)
虚数化简i^(7x+2)
还有i^(6x-1),i^(14x-11)
1回答
2020-09-23 01:56
虚数化简i^(7x+2)还有i^(6x-1),i^(14x-11)
虚数化简i^(7x+2)
还有i^(6x-1),i^(14x-11)
i^(6x-1)=i^6x/i=(i^2)^3/i=-i^(x-1)
i^(14x-11)=i^14x/i^11=(i^2)^7/-i=i^(x-1)