If 1+i1−ix=1 , then

1+i1−ix=(1+i)(1+i)(1−i)(1+i)x =(1+i)21−i2x=1−1+2i2x⇒1+i1−ix=(i)x=1  (given)⇒   (i)x=(i)4n ,Where n  is any positive integer⇒   x=4nNote: 4 is the least value for which i4=1 .