00:01
So in first part of the question, we are given that integration of i equals to integration from c r, ez, pz.
00:08
So the path of the integration p, so z will be equals to 0 to z equals to 1 plus 2 iota.
00:18
So as z equals to x plus iota y, so from where we will get that r, ez equals to x and pz equals to dx plus iota dy.
00:33
So now equation of c will be y equals to 2x which is a line.
00:43
So therefore, dz will be equals to dx plus iota d2x as y equals to 2x.
00:53
So from where we will get dz equals to dx plus 2 iota dx.
01:01
So this will be equals to 1 plus 2 iota dx.
01:07
Therefore, we get that integration of x, 1 plus 2 iota dx integration from x equals to 0 to 1 of 1 plus 2 iota x dx.
01:26
So from where we will get that integration i equals to 1 plus 2 iota x square divided by 2 from 0 to 1 and this will be equals to 1 plus 2 iota...