 Hi and welcome to the session. I am Asha and I am going to help you with the following question which says find the modulus of 1 plus iota upon 1 minus iota minus 1 minus iota upon 1 plus iota. For any complex number z where z is equal to 8 plus iota b then modulus of z which is denoted by this sign is equal to root over a square plus b square. So with the help of this formula we will find the modulus of this complex number. So this is our key idea. Let us now begin with the solution and here the given complex number is 1 plus i upon 1 minus i minus 1 minus i upon 1 plus i or it can further be written as 1 minus i into 1 plus i in the denominator and the numerator we have 1 plus i whole square minus 1 minus i whole square or 1 plus iota square plus 2 iota minus 1 plus 2 iota minus iota square upon in the denominator. Flying the formula of a minus b into a plus b we have 1 square minus iota square since a minus b into a plus b is equal to a square minus b square or it can further be written as 1 plus iota square is minus 1 plus 2 iota minus 1 plus 2 iota minus of minus 1 upon 1 minus iota square is minus 1 therefore it can further be written as 1 minus 1 plus 2 iota minus 1 plus 2 iota plus 1 and in the denominator we have 1 plus 1 or now cancelling 100 minus 1 this minus 100 plus 1 we have 4 iota upon 2. Now taking modulus of the denominator and numerator root upon 2 modulus which is equal to modulus 4 iota upon modulus 2 which is further equal to root over 4 square and in the denominator root over 2 square since we know that f z is equal to a plus iota b is any complex number and modulus of z is equal to root over a square plus b square and here we have a equal to 0 and b is 4 so b square is 4 square and in the denominator b is 0 and a is 2 so a square is 4 which is simplifying comes equal to 4 upon 2 which is equal to 2 thus we can say that our answer is modulus of z is equal to 2. So this completes the solution hope you enjoyed it have a good day.