 I am welcome to the session, I am Arshaun and I am going to help you with the following question which says if x minus iota y is equal to a minus iota b upon c minus iota d, prove that x square plus y square is equal to a square plus b square upon c square plus d square. So, let us begin with the solution and we have given that x minus iota y is equal to a minus iota b upon c minus iota d. Now, removing i from the denominator. So, let us rationalize it by multiplying the numerator and denominator by c plus iota d. On the denominator we will have c square minus iota d whole square since a minus b into a plus b is equal to a square minus b square and here in place of a we have c and in place of b we have iota d. So, like this formula and in the numerator we have a minus iota b into c plus iota d. Now, it can be further written as a c plus iota ad minus iota bc minus iota square bd upon in the denominator c square minus iota square d square. Now, since iota square is equal to minus 1 therefore, this can further be written as a c plus bd since iota square is minus 1 and minus into minus is plus to plus dd and now taking i common from these two terms we have ad minus bc and in the denominator we have c square plus d square which can further be written as by separating the real parts and imaginary parts a c plus bd upon c square plus d square plus iota times of ad minus bc upon c square plus d square and this is equal to x plus iota y. Now on comparing the real and imaginary parts we find here that x is equal to a c plus bd upon c square plus d square and y is equal to ad minus bc upon c square plus d square with a negative sign since here we have minus iota and here we have plus iota so we have taken minus sign common and thus y is equal to bc minus ad upon c square plus d square. Now, let us find out value of x square plus y square now x is ac plus bd upon c square plus d square whole square and y is bc minus ad upon c square plus d square whole square. Now looking the squares we have a square c square plus b square d square plus 2 times of a bcd and in the denominator we have c square plus d square whole square. Similarly, opening this bracket we have in the numerator b square c square plus a square d square minus 2 times of a bcd and in the denominator we have c square plus d square whole square. Now let us take LCM so in the denominator we will have c square plus d square whole square and in the numerator we will have a square c square plus b square d square plus 2 times a bcd plus b square c square plus a square d square minus 2 times of a bcd. Now let us cancel light terms with unlike signs and we are left with a square c square plus b square d square plus b square c square plus a square d square whole c square plus d square whole square and now taking a square common from these 2 terms here c square plus d square and taking b square common from these 2 terms we have b square into c square plus d square upon c square plus d square whole square. Now again taking c square plus c square common we have d square plus b square and in the denominator we have c square plus d square whole square. Now the c square d square cancels out with 1 c square d square in the denominator and we have a square plus b square upon c square plus d square. Hence we have proved that x square plus y square is equal to b square plus b square plus b square upon c square plus d square. So this completes the solution hope you enjoyed it take care and have a good day.