CSIR NET MATHEMATICS | Complex Analysis | Concept & Solution Complex Integration June 2019

Hello friends, in this video we are going to discuss the problem of complex analysis from CSI NET, June 2019. Example number 33, Bouclier Seam Marks 3. Given a real number a greater than 0, consider the triangle delta with vertices 0, a, a plus i in. If delta is given the counterclockwise orientation, then the contour integral integration over Over delta REZDZ with REZ denoting the real part of Z. is equal to first option 0 second option i a square by 2 third option i a square and the fourth option i 3 a square by 2. so it is a problem of complex integration so we can with the region which is triangle with vertices 0 0 second a so capital a a and the third what is a plus i a so which is b a plus i a And the orientation is counterclockwise. Now, z is equal to x plus iy. Real part of z is x. And dz is equal to x plus idy. Our integration is integration over triangle. Real part of z dz. So, your real part is x. So, x and dz is equal to dx plus idy. Now, this triangle. The region is a non-continuous region. So, we divide into the smaller continuous part. And we have three continuous part on the boundary OA, AB and BO. So we have three part of the integration this integration over triangle is equal to integration over OA plus integration over AB plus integration over BO line segments. So we begin with the integration over the line segment OA. OA is a part of x-axis so y is equal to 0 it's a equation of OA line so dy is equal to 0 and x vary from 0 to a now substitute this value then we have integration 0 to a x dx and dy equal to 0 now by simply integration we have a square by 2 so integration over line segment OA as value a square by 2 now similarly So, integration over the line segment AB. So, AB is a vertical line and its equation is X is equal to A because its X coordinate is fixed and so the DX is equal to 0 and value of Y vary from 0 to A. Now substitute this value in the integration. We have integration 0 to A, X is equal to A, DX is equal to 0 and DY equal to 0. So, we have integration 0 to A, A, I, D, Y. By simply integration, I, A constant, so 1 D, Y, its integration is Y, evaluate between 0 to A, so it is A, so I, A square. So, integration over line segment A, B is I, A square. Now, the third line segment, B, O, so its line equation is X is equal to Y, then D, X is equal to D, Y. And x vary from a to 0 because it is counter clause by direction line segment begin with a and end with the 0. So x has limit a to 0. Now substitute the value. So integration a to 0 x dx plus i dy is equal to dx. X 1 plus i into dx. So 1 plus i is constant. Take out from the integration. Integration of x is x square by 2. Evaluate between the limits. a to 0 then we have upper limit is 0 and the lower limit is a so we have minus a square by 2. So over the line segment b o value is 1 plus i minus a square by 2. Now by summing the all the three result of integration a square by 2 plus i square plus 1 plus i minus a square by 2. After the simplification Minus a square by 2 cancel with a square by 2 and i square minus i square by 2 we have i square by 2. So this integration has value i square by 2 and option is 2. So second option is the correct answer. Thanks for watching. For more videos like share subscribe. Subscribe. Place the bell icon of next gen mass.