0 votes 0 votes Rajat Sharma 1 asked Dec 15, 2015 Rajat Sharma 1 6.3k views answer comment Share Follow See 1 comment See all 1 1 comment reply l0k3ndr commented Oct 3, 2019 reply Follow Share I don't think it is wrt multiplication. As pointed by others, 0 won't have a inverse in this which could lead to identity element of 1. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes yes , complex number under multiplication forms group. for example It is closed Associativity property satisfied for any element identity element is 1 =w^3 inverse of 1 is 1, w is w^2, w^2 is w So, it will be a group under multiplication srestha answered Dec 16, 2015 edited Dec 16, 2015 by srestha srestha comment Share Follow See all 4 Comments See all 4 4 Comments reply Rajat Sharma 1 commented Dec 16, 2015 reply Follow Share you took a particular instance?I want to know whether it is true for all instances? 1 votes 1 votes srestha commented Dec 16, 2015 reply Follow Share yes it is applicable for all 0 votes 0 votes Rajat Sharma 1 commented Dec 16, 2015 reply Follow Share I got to know that 0 is a complex number.How will you calculate its inverse in a set {0, all other complex numbers}? 1 votes 1 votes Gaurav Yadav commented Sep 10, 2019 reply Follow Share Additive and multiplicative inverse exist only for non-zero elements. Additive inverse is -a-ib and multiplicative inverse is 1/(a+ib) = a/(a^2+b^2)+i(-b/(a^2+b^2). But there does not exist any inverse for (0+i0). Therefore, set of all complex numbers under multiplication is not a group. 0 votes 0 votes Please log in or register to add a comment.