 Hi and welcome to the session. I am Arsha and I am going to help you with the following question which says convert the following in the polar form and we have to convert 1 plus 7 out of 2 minus iota whole square. Now as we know standard form of the complex number is a plus iota b and the polar form it can be written as z is equal to r cos theta plus iota sin theta rs the square root of x square plus y square which is also called the modulus of z theta is called the argument of z. So these are some ideas which we are going to use to convert the given into its polar form so they are our key ideas. Let us now begin with the solution and we have 1 plus 7 iota upon 2 minus iota square. Now there is simplified further numerator as it is and I am opening the square and the denominator we have 4 minus 4 iota plus iota square since a minus b whole square is equal to a square minus 2ab plus b square and here in case of a we have 2 and in case of b we have i thus we can write it further as 1 plus 7 iota upon. Now iota square is minus 1 and 4 minus 1 is 3 so we have 3 minus 4 iota under denominator. Now let us rationalize it to remove i from the denominator so multiplying the numerator and denominator by 3 plus 4 iota which is the conjugate of 3 minus 4 iota we have in the denominator 3 square minus 4 iota whole square since a minus b into a plus b is equal to a square minus b square and here in case of a we have 3 and in case of b we have 4 iota and in the numerator we have 1 plus 7 iota into 3 plus 4 iota. Now let us simplify the numerator first 3 plus 4 iota plus 21 iota plus 28 iota square and in the denominator we have minus 9 minus 16 iota square just further equal to now iota square is minus 1 so this becomes minus 28 and minus 28 plus 3 is minus 25 plus 25 iota which we get on adding 4 iota plus 21 iota and in the denominator we have 9 minus 16 into minus 1 since iota square is minus 1 so we have taking 25 common from the numerator minus 1 plus iota upon 25 and now canceling the common factor we have minus 1 plus iota. Now let x plus iota y is equal to minus 1 plus iota and on comparing we find that here x is equal to minus 1 and y is equal to 1 so first let us find r which is root over x square plus y square so this implies root over minus 1 square plus 1 square which is equal to root over 2. In the polar form any complex number can be written as r cos theta plus iota sin theta if we have z is equal to x plus iota y which is minus 1 plus iota so we have minus 1 plus iota is equal to r cos theta plus iota sin theta and on comparing this we find that r cos theta is equal to minus 1 and sin theta is equal to 1 and r is root over 2 so this implies cos theta is minus 1 upon root over 2 and sin theta is 1 upon root over 2 which further implies that theta is equal to 3 pi by 4 which is the argument of z the complex number and the polar form can be written as root over 2 which is the value of r into cos theta theta is 3 pi by 4 plus iota sin theta which is again 3 pi upon 4 so this is the required polar form of the given expression so this completes the solution hope you enjoyed it take care and have a good day.