if S(n)=i^+i-^where i=√-1 and ^=n is a positive integer then find the total number of distinct value of S(n)??????? Share with your friends Share 0 Vishvesh Kumar answered this Given that S(n)=in+i-n where i=-1 and n is positive integer Note that any positive integer n is one of the following form 4k, 4k+1,4k+2 and 4k+3 where k is any non negetive intergerWe know that i2=-1 and i4=1and hence note that i4k=(i4)k=(1)k=1 and so i4k+1=i ,i4k+2=-1 ,i4k+3=-iHence S(4k)= 1+1 =2 S(4k+1) =i+i-1= i+i=2i { i-1=i } S(4k+2) =-1+(-1)-1=-1+-1=-2 S(4k+3) =-i+(-i)-1=-i-(i)-1=-i-i=-2iSo we get S(n)= ±2 ,±2i four different values 0 View Full Answer
