 Hi and welcome to the session, I am Asha and I am going to help you with the following question that says, find the modulus and argument of the complex number 1 plus 2 iota upon 1 minus 2 iota. Z is any complex number if we form X plus iota Y in the polar form Z can be written as cos theta plus iota sin theta where it is equal to the modulus of Z is equal to root over X square plus Y square and this theta is called the argument of Z it is also denoted by A R G Z. So, these are some ideas with the help of which you will find the modulus and argument of the given complex number. So, these are the ideas. Let us now begin with the solution let the given complex number be denoted by Z. So, Z is equal to 1 plus 2 i upon 1 minus 3 iota. Now, multiplying the numerator and denominator by the conjugate of 1 minus 3 iota which is 1 plus 3 iota thus this can further be written as in the denominator we have 1 square minus 3 iota whole square since A minus B into A plus B is equal to A square minus B square and here A is 1 and B is 3 iota and in the numerator we have 1 plus 2 iota into 1 plus 3 iota which can further be written as 1 plus 3 iota plus 2 iota plus 6 iota square upon 1 minus 9 iota square. Now, since iota square is equal to minus 1 therefore, it can further be reduced to this gives minus 6 and minus 6 plus 1 is minus 5 plus on adding 3 iota with 2 iota we have 5 iota and in the denominator this gives minus 9 and minus into minus is plus so in the denominator we have 10. Now, taking 5 common from the numerator we have minus 1 plus iota upon 10 and now canceling the common multiple the numerator and denominator we are left with minus half plus iota upon 2. Now, the standard form of a complex number is x plus iota y and here z is equal to minus half plus iota into half. So, in comparing we find here that x is equal to minus half and y is equal to half. Firstly, let us find the value of modulus of z which is root over x square plus y square on substituting the values we have minus half square plus half square which is equal to root over 1 upon 4 plus 1 upon 4 which is further equal to root over 2 upon 4 which further gets 1 upon root over 2 thus modulus of z is equal to 1 upon root over 2. Now, in the polar form z can be written as r into cos theta plus iota sin theta where r is the modulus of z which is equal to 1 upon root 2. Therefore, we can write it as x plus iota y is equal to 1 upon root 2 cos theta plus iota sin theta and x plus iota y is minus half plus half iota which is equal to 1 upon root 2 cos theta plus iota sin theta and now on comparing the real and imaginary parts we have minus half is equal to 1 upon root 2 cos theta half is equal to 1 upon root 2 sin theta this further implies that cos theta is equal to minus 1 upon root 2 and sin theta is equal to 1 upon root 2 and this implies theta is equal to 3 pi by 4 which is the argument of z. So, the answers are modulus of z is 1 upon root over 2 and argument of z is 3 pi upon 4. So, this completes the solution hope you enjoyed it take care and have a good day.