00:01
Hello students, we need to evaluate the following using residue theorem.
00:05
Given integral over the boundary c 3z plus 2 whole square divided by z into z minus 1 into 2z plus 5 dz.
00:19
Given the condition mod z, that is boundary c, is equal to 3.
00:25
Now here we can observe that we have alpha is equal to 0, 1 and minus 5 by 0.
00:32
Simple poles of f of z is given by that is f of z we have the value 3 z plus two whole square divided by z into z minus 1 into 2 z plus 5 now residue of z comma z is equal to 0 we have limit z 10 into 0 into z minus 0 into f of z this implies we have limit z ending to 0 3 z plus 2 whole square divided by z into z minus 1 into 2 z plus 5.
01:16
This implies we have 4 by minus 1 into 5 which gives the value minus 4 by 5.
01:23
Now residue of f of z that is z is equal to 1 we have 3 into 1 plus 2 whole square divided by 1 1 1 1 2 into 1 plus 5 which gives the value 25 by 7 and residue of f of z when z is equal to minus 5 by 2 we have minus 15 by 2 plus 2 whole square divided by minus 5 by 2 into minus 5 by 2 minus 5 by 2 minus 1 which gives the value 12 by 21 by 35 therefore integral over the boundary we see f z into d z is equal to 2 pi i into summation of residue of f of z into where z belongs to alpha.
02:18
This implies we have 2 pi i into minus 4 by 5 plus 25 by 7 plus 121 by 35.
02:29
On calculating and simplifying, therefore we get the value 436 pi i divided by 35...
