 Hello and welcome to this session. In this session we will discuss a equation which says that find the modulus of 6 plus 3 iota whole square whole upon 3 minus 2 iota. Now before starting the solution of this equation, we should know our result. And that is the absolute value modulus 2 complex number z such that z is equal to a plus b iota where a and b belongs to the set of real numbers is given by modulus of z is equal to modulus of a plus b iota which is equal to square root of a square plus b square. Now the modulus of this number is just the distance from the region to this number. Now this result will work out as a key idea to this question and now we will start with the solution. Here we have to find the modulus of this. So given iota whole square whole upon 3 minus 2 iota. Now in the numerator apply the formula of a plus b whole square which is equal to a square plus b square plus 2 a b. So this will be equal to 6 square iota whole square plus 2 into 6 into 3 iota all upon 3 minus 2 iota. Now this is equal to 36 plus 9 iota square plus minus 2 iota plus 9 into iota square is equal to minus 1 so it will be into minus 1 plus 36 iota whole upon iota. Now this will be equal to 36 minus 9 plus 36 iota whole upon whole square whole upon iota is equal to 36 minus 9 is 27 iota whole upon 3 minus 2 iota. 3 iota whole square whole upon 3 minus 2 iota is equal to, now to rationalize this we will multiply the numerator and the denominator by the conjugate of this. The conjugate of this is 3 plus 2 iota. So this will be equal to iota whole upon 3 minus 2 iota into whole upon 3 plus 2 iota. Now this will be equal to 27 plus 36 iota whole into 3 plus 2 iota whole into 3 plus 2 iota whole 81 iota plus 54 iota into 3 is 108 iota plus 2 iota equal upon. Now here in the denominator we will apply the formula of A minus B the whole into A plus B the whole which is equal to A square minus B square. Now A here is 3 and B here is 2 iota. So in the denominator it will be 3 square however this is equal to 81 plus 162 iota plus 72 into iota square is minus 1 so it will be minus 1 here. Whole upon 3 square is 9 plus 162 iota minus 72 whole upon 9 minus 4 into iota square is minus 1 72 is 9 plus 162 iota whole upon 9 minus 4 into minus 1 will be plus 4. So it will be 9 plus 4 which is equal to 13 whole upon is equal to 9 plus 162 iota whole upon 13 which can be written as 9 over 13 plus 162 by 13 iota. Now using this result which is given in the key idea now here B equal to 162 by 13. Now the modulus of iota whole square iota is equal to square root of so this will be 9 by 13 whole square. Now this is equal to square root by 169 to 244 69 which will be equal to square root of now here by taking the hm which is 169 here it will be 81 plus 44 which is equal to 26,335. Now here 13 into 13 is 169 and 13 into 2025 and 25. So this is equal to 2025 over 13. Now this is equal to root 2025 root 13 which is equal to root 30 45. Iota whole square whole upon iota is equal to by root 13. This is another given question and that's all for this session. Hope you all have enjoyed this session.