 Hello and welcome to the session. In this session we will learn about symmetrical interpretation of triangular inequality. Now the triangle inequality is modulus of Z1 plus Z2 is less than equal to modulus of Z1 plus modulus of Z2 where Z1 and Z2 are two complex numbers. Now let us discuss the symmetrical interpretation of triangular inequality which is modulus of Z1 plus Z2 is less than equal to modulus of Z1 plus modulus of Z2. Now let the complex number Z1 is equal to X1 plus Y1 iota and the complex number Z2 is equal to X2 plus Y2 iota be represented by the points that is Z1 is represented by the point and Z2 is represented by the point Q in the argument diagram in the complex plane that is if the point Q represents the complex number Z1 then the coordinates of point P are X1 Y1 and if the point Q is representing the number Z2 that is the complex number Z2 then the coordinates of point Q are X1 Y2 so we have drawn the points P and Q on the argument plane. Now join the original O the points P then complete the parallelogram coordinate P RQ so this is the parallelogram Q with OP and OQ as the adjacent sides. Now draw the perpendicular to the X axis perpendicular to so now we have drawn the perpendicular to Q is equal to angle P L R 90 degrees and the sides OQ is equal to these are the opposite sides the parallelogram and opposite sides of the parallelogram are parallel and equal and let this be angle 1 and this be angle 2 and angle 2. Now let us take this angle 3 and this as angle 4 now Q L perpendicular to the line that is the X axis that means Q L is parallel to the transversal is equal to 180 degrees these will be the interior right angle the parallelogram that means equal to 180 degrees these are also interior right angles so from 1 is equal to angle that is equal to angle 3 can be written as angle 2 that is this angle angle 2 plus angle 4 that is this complete angle can be written as angle 1 plus angle 3 angle 4 now this implies angle 1 is equal to and 1 is equal to angle 2 this we have 2 therefore by angle angle side property the angle is equal to P L that is corresponding parts of congruent triangles are congruent. Now from the diagram you can see that P L is equal to M S so we have from this angle is equal to P L is equal to M S as we have drawn a point Q whose coordinates are X 2 Y 2 that is this point is at a distance on the region and is equal to X 2 the Y coordinate of Q is Y 2 that means Q L is equal to Y 2. Now O S is equal to X 1 that means is equal to X 1 and we have proved M S is equal to X 2 so this is equal to is equal to Y 1 as the Y coordinate of P is Y 1 therefore S L is also Y 1 and we have proved R L is equal to Y 2 S which is equal to Y 2 plus Y 1 which is equal to Y 1 plus Y 2 S is equal to X 1 plus X 2 Y 1 plus Y 2 that is the X coordinate of R is X 1 plus X 2 and Y coordinate of R is Y 1 plus Y 2 the coordinates the complex number plus Y 2 the whole into Iota Z 1 plus Z 2 Z 1 is equal to X 1 plus Y 1 Iota and Z 2 is equal to X 2 is equal to Iota and there is some that is Z 1 plus Z 2 is equal to X 1 plus X 2 the whole plus Y 1 plus Y 2 the whole into Iota now from the triangle is less than OP plus P R because in a triangle is greater than the third side the complex number to the point that is representing the complex number so here the distance OP is the complex number Z 1 as the point P is representing the complex number Z 1 and opens of the original the modulus of Z 1 there is equal to O Q as these are the opposite sides of the parallelogram the number that is the complex number Z 2 that means O Q is equal to the modulus of Z is equal to O Q so P R will be equal to the modulus of Z 2 the complex number Z 1 plus the modulus of Z 1 plus Z 2 Z 1 plus Z 2 that is the distance O R is less than OP which is the modulus of Z 1 is the modulus of now let us name this in equation as R lying on the same line OP plus P R which implies the modulus of Z 2 is equal to modulus of Z 1 plus modulus of surplus Z 2 is less than equal to modulus of Z 1 nin modulus of on example Z 1 is equal to 2 plus 3 IOTA that is the complex number Z 1 is equal to 2 plus 3 IOTA the complex number Z 2 is equal to minus 2 minus per IOTA is further equal to 0 modulus of Z 1 plus Z 2 minus which is equal to minus 1 square and this is equal to 1 and modulus of Z1 will be equal to square root of t square which is equal to root which is equal to root 13. And modulus of Z2 will be equal to square root of minus 2j 바로 square which is equal to which is equal to root 20 which is further equal to root 5. of z1 plus z2 is less than modulus of z1, z1 plus z2 is less than modulus of z1 plus modulus of z2. In the second case, here let the complex number z1 is equal to 1 plus 2 iota by z2 is equal to 2 plus then z1 to the whole plus 2 plus 4 iota to the whole which is further equal to 3 plus z1 plus z2 is equal to square root of a square plus b square which is square root of 3 square plus 6 square which is equal to root 9 plus 36 which is equal to root 45 and which is further equal to 3 root 4 of z1 is equal to square root of 1 square plus 2 square which is equal to square root of 1 modulus of z2 is equal to square root of which is equal to square root is equal to root 20 and this can be written as modulus of z1 plus modulus of z2 is equal to root 5 which is equal to 3 root 5. Now modulus of z1 plus z2 is equal to 3 root 5, modulus of z1 plus modulus of z2 is equal to 3 root 5. Therefore, z1 plus z2 is equal to modulus of z1 plus modulus of z2. Now from these examples we can observe that either modulus of z1 plus z2 is less than modulus of z1 plus modulus of z2. Modulus of z1 plus z2 is equal to modulus of z1 plus modulus of z2. Modulus of z1 plus z2 is less than equal to modulus of z1 plus modulus of z1 plus modulus of z1 plus modulus of z1 plus modulus of z2. The triangle inequality we have We have learnt about the symmetrical intrepitation of triangular inequality. Enjoy the session.