Please tell how to solve question 7.Asap Share with your friends Share 0 Ankita Agarwal answered this Dear Student , Consider 1+in1-in-2Rewrite it as1+in1-in-2=1+i1-in-2×1+i2 =1+i×1+i1-i1+in-2×1+i2 =1+i212-i2n-2×1+i2Using a+b2=a2+b2+2ab , a-ba-b=a2-b2 =1+i2+2i1--1n-21+i2+2i i2+1=0 =2i2n-22i =in-2×2i =2in-1 if n=1 =2i1-1 =2Then the expression is real number,Thus the smallest positive integer value of n is 1 Regards 0 View Full Answer
