 Hi and welcome to the session. I am Asha and I am going to help you with the following questions. This says reduce 1 upon 1 minus 4 iota minus 2 upon 1 plus iota into 3 minus 4 iota upon 5 plus iota to the standard form. Now the standard form of a complex number is of the A plus iota B. So we will try to reduce this in the form of A plus iota B. So let us begin with the solution and we are given 1 upon 1 minus 4 iota minus 2 upon 1 plus iota into 3 minus 4 iota upon 5 plus iota which can further be written as Lcm of 1 minus 4 iota and 1 plus iota is their product. The numerator we will have 1 plus iota minus 2 times of 1 minus 4 iota into 3 minus 4 iota upon 5 plus iota. It is further equal to 1 plus iota minus 2 plus 8 iota upon in the denominator we have minus 3 iota plus 4 into 3 minus 4 iota upon 5 plus iota and let us solve first 1 minus 4 iota into 1 plus iota to show that the denominator is equal to 1 minus 3 iota plus 4. Now this will be equal to 1 plus iota minus 4 iota and minus into plus is minus so minus 4 iota square and since iota square is equal to minus 1 therefore this will become 1 plus 4 and on simplifying this we get minus 3 iota. So we have 1 minus 3 iota plus 4 which is further equal to the numerator we have minus 1 plus 9i upon in the denominator we will have 5 minus 3 iota into 3 minus 4 iota upon 5 plus iota as a test which can further be written as minus 1 plus 9 iota into 3 minus 4 iota upon 5 minus 3 iota into 5 plus iota. Now on simplifying the numerator and denominator we have minus 3 plus 4 iota plus 27 iota minus 36 iota square and in the denominator we have 25 plus 5 iota minus 15 iota minus 3 iota square which is further equal to minus 3 27 iota plus 4 iota is 31 iota minus 36 iota square is minus 1 upon in the denominator we have 25 on simplifying 5 iota minus 15 iota we have minus 10 iota minus 3 into iota square is again minus 1 which can further be written as minus 3 plus 31 iota plus 36 on in the denominator we have 25 minus 10 iota plus 3 which can further be written as 33 plus 31 iota the denominator 28 minus 10 iota and now let us rationalize it so we have 33 plus 31 iota upon 28 minus 10 iota into 28 plus 10 iota upon 28 plus 10 iota. Now in the denominator we have 28 square minus 10 iota square since A minus B into A plus B is equal to A square minus B square and every place of A we have 28 and in place of B we have 10 i and in the numerator we have 33 plus 31 iota into 28 plus 10 iota now let us simplify the numerator first we have to multiply 33 by 28 plus 33 into 10 iota plus 31 iota into 28 plus 31 iota into 10 iota and in the denominator 28 square is 784 minus 10 iota square is 100 iota square which is further equal to 924 plus 330 iota plus on multiplying 31 iota with 28 we get 868 iota plus multiplying 31 iota with 10 iota we have 310 iota square and in the denominator we have 784 minus 100 iota square is minus 1 so it is further equal to 924 on adding 330 iota with 868 iota we get 1198 iota plus 310 and iota square is minus 1 and in the denominator we have 784 plus 100 which is further equal to now 924 minus 310 is 614 plus 1198 iota the denominator we have 884 now taking two common from the numerator we have 307 plus 599 iota on the denominator we have 884 now I am cancelling the common factors of numerator and denominator we have 307 plus 599 iota upon 442 so on reducing the given problem we get our answer as 307 plus 599 iota upon 442 so this completes the solution hope you enjoyed it take care and have a good day.