 Hi and welcome to the session, I am Asha and I am going to help you with the following question that says, if i plus 1 upon 1 minus ieta raised to the power m is equal to 1, then find the least integral value of m. Now as we know, iota square is equal to minus 1. This implies iota q is equal to iota square into iota which is equal to minus iota. Similarly iota 4 can be written as iota square into iota square which is equal to minus 1 into minus 1. This is equal to 1. Hence, we can say that iota raised to the power 4 is equal to 1. By using these few ideas, we will try to find the integral value of m for which this is satisfied. So this is a key idea. Let us now begin with the solution and here we are given 1 plus iota upon 1 minus iota raised to the power m is equal to 1. So we have 1 plus iota upon 1 minus iota and we will try to write that in the form for complex number a plus iota b, so multiplying the numerator and denominator by the conjugate of 1 minus iota which is 1 plus iota, we have in the denominator 1 square minus iota square and in the numerator we have 1 plus iota whole square which is further equal to 1 plus 2 iota plus iota square and in the denominator we have 1 square minus of minus 1. Since iota square is equal to minus 1, therefore in the numerator we have 2 iota in the denominator we have 2. Since here again iota square is minus 1, so minus 1 plus 1 cancels out and we are left with i. So we have 1 plus iota upon 1 minus iota is equal to i and we are given that 1 plus iota upon 1 minus iota raised to the power m is equal to 1, this implies i raised to the power m is equal to 1 and we have to find the least integral value of m and by a key idea we know that i raised to the power 4 is equal to 1 this implies m is equal to 4 that is the least integral value of m is equal to 4. So this completes the solution, hope you enjoyed it, take care and have a good day.