 Hello and welcome to the session. In this session we will discuss a question which says that find the argument of the given complex numbers. First is z1 is equal to 403 plus 4 iota. Second part z2 is equal to minus 403 plus 4 iota. Third part is z3 is equal to minus 403 minus 4 iota. Now before starting the solution of this question we should know our result and that is our complex number z which is equal to a plus b iota. The argument of z an angle that is of z is equal to theta which is equal to tan inverse modulus of b by a. Now this point b is representing the number that is the complex number z and here this is the angle theta with initial size the positive x axis and the terminal side they will form the origin that is this point of z that is the complex number z and the theta will be equal to tan inverse modulus of b by a. Now suppose we have to find the value of tan inverse modulus of b by a line between 0 and pi by 2 and let it be the angle alpha. Now let the complex number z is equal to a plus b iota. Now let this complex number is represented by the point p whose coordinates are ab. Now let us find out the quadrant in which this number pab lies. Now making this table in which we will discuss the various values of a and b then we will check in which quadrant this point pab lies and then we will check the argument of z. The first case is when a is greater than 0 check greater than 0 then in that case quadrant and in this case argument of z will be equal to alpha. Now let us discuss the second case in this greater than 0 then this point will lie in the second quadrant and in this argument of z will be equal to is less than 0 and b is also less than 0. Then in that case in the third quadrant the second of z will be equal to minus now in the next case greater than 0 and b is less than 0 then this point will lie in the second of z will be equal to minus alpha. Now let us start with the first part. Now given the complex number z1 is equal to 4 root 3 and inverse the modulus of b by a. Now where a is 4 root 3 and b is alpha will be equal to 10 inverse modulus of b by a. Which is equal to 10 inverse modulus of 4 by 4 root 3. Now 10 inverse 1 by root 3 is equal to pi by 6 therefore alpha is equal to pi by 6. b that means a is greater than 0 that means b is greater than 0. So here a is greater than 0. Now using this result which is given in the key idea b is greater than 0 lies z1 is equal to alpha which is equal to pi by 6. Now in the second part the complex number z2 is given as minus 4 root 3 plus 4i. Now here a is equal to minus 4 root 3 therefore alpha is equal to 10 inverse modulus of b by a which will be equal to 10 inverse modulus of minus 4 root 3. Since I am taking the modulus it will be equal to 10 inverse 1 by root 3 and 10 inverse 1 by root 3 equal to minus 4 root 3 is greater than 0. Now using this result which is given in the key idea is less than 0 and b is greater than 0. So the point a b which is here of z2 is equal to pi minus 0 to pi minus pi by 6. Now in the third part z3 is given as minus 4 root 3. Now here a is equal to minus 4 root 3 and b is equal to minus 4. So alpha will be equal to 10 inverse modulus of b by a which will be equal to 10 inverse modulus of minus 4 by minus 4 root 3 and this will be equal to I am solving and taking the modulus it will be 10 inverse 1 by root 3 which is equal to 10 is equal to pi by 6. So alpha will be equal to minus 4 root 3 which is less than 0 which is also less than 0. Now using this result which is given in the key idea is less than 0 b is less than 0 therefore the point a b that is the point minus 4 root 3 minus of z3 will be equal to minus 1 by root 3 which will be equal to minus alpha by 6. The whole this is equal to minus z3 is equal to which will be equal to pi by 6 which I am solving is equal to 7 pi by 6 and the z4 is equal to 4 root 3 minus 4 I am solving is equal to minus 4. Alpha will be equal to 10 inverse the modulus of b over a which will be equal to minus 4 over 4 root 3 and this is equal to 10 inverse modulus of minus 1 by root 3 which is equal to 10 inverse 1 by root 3 is equal to pi by 6 is greater than 0 and b is equal to minus 4 which is less than 0. Now using this result which is given in the key idea as 0 and b is less than 0 and therefore z4 is equal to minus which will be equal to minus pi by 6 root is equal to 2 pi minus alpha which will be equal to pi by 6 which I am solving is equal to 11 pi. That is the solution of the given question and that is all for this session. Hope you all have enjoyed this session.