 Hello and welcome to the session. In this session we will discuss a question which says that if z1 is equal to 4 minus 2 iota and z2 is equal to 3 minus 6 iota, plot the number z1 plus z2 also show that modulus of z1 plus modulus of z2 is greater than modulus of z1 plus z2. Now before starting the solution of this question we should know what is out. And that is modulus of any complex number z which is equal to a plus b iota where a and b belong 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 where the modulus of a plus b iota is just the distance from the region in the client a plus b iota. Now this result will work out as a key idea for solving this question. And now we will start with the solution. Now z1 and z2 are going to us and we have to plot this number. So given z1 is equal to 4 minus 2 iota and z2 is equal to 2 minus 6 iota. Therefore z1 plus z2 is equal to 4 minus 2 iota Rahul plus 2 minus 6 iota Rahul. And solving will be equal to 4 plus 2 will be 3 minus 6 iota will be minus 8 iota. Now z1 which is equal to 4 minus 2 iota is represented by the point a whose coordinates are 4 minus 2. Also z2 which is equal to 2 minus 6 iota is represented by the point b whose coordinates are 2 minus 6. This is z1 plus z2 which is equal to 6 minus 8 iota is represented by the point c whose coordinates are 6 minus 8. These points on the graph point a whose coordinates are 4 minus 2 the number z1. Now we will plot the point b whose coordinates are point b which is represented by the number z2. This number that is the complex number z1 plus z2 on the graph and it is represented by the point c whose coordinates are 6 minus 8 which is representing the number z1 plus z2. So we have plotted the number z1 plus z2. First we will use the z which is given in the key idea. Now z1 is equal to 4 minus 2. So modulus of z1 will be equal to square root of a square which is 4 square plus b square which will be minus 2 square. So this is equal to square root of 16 plus 4 which is equal to root 20 which is equal to 2 root 5. Now modulus of z1 is just the distance from the region to this point. Therefore modulus of z1 will be equal to the distance of a. So modulus of z1 the distance of 8. This of z2 will be equal to a square which is 2 square which is minus 6 square. This is equal to square root of 4 plus 36 which will be equal to 2 root 10. Now again modulus of z2 this point is from the region. Therefore modulus of z2 is equal to the distance of b. So this is equal to the distance of b. Now z1 plus z2 is equal to 6 minus 8 iota. Therefore modulus of z1 plus z2 is equal to square root of 6 square plus b square which is minus 8 square. Which is equal to square root of 36 plus 64 which is equal to root 100 which is equal to 10. The distance from the region is equal to the distance of. So this is the value of modulus of z1. This is modulus of z2 is greater than is equal to the distance of b. Therefore it is greater than modulus of z1 plus the distance of b which is modulus of z2 is greater than the distance of b which is modulus of z1 plus the attrition of the given question. And that's all for this session. Hope you all have enjoyed this session.