00:01
In this question we have this function fz is equal to z divided by z minus 1 into z plus 1 whole square and we have to find out the residues at each of its poles.
00:19
So first of all let us find out poles for this function.
00:23
Here clearly we can see that z is analytic at 0.
00:27
So for poles of fz we will put the denominator that is z minus 1 into z plus 1 whole square equal to 0.
00:36
So from here z minus 1 equal to 0 means z is equal to 1 and z plus 1 equal to 0 means z is equal to minus 1.
00:46
Here this is the simple pole since it is of order 1 and here we have order 2 so z equal to minus 1 will be of order 2.
00:59
So we have these two poles z equal to 1 and z equal to minus 1.
01:04
Let us find the residues at each pole.
01:08
So for z is equal to 1 the residue will be equal to limit z tending to 1 z minus 1 into fz.
01:22
That means this will be limit z tending to 1 z minus 1 into fz is z over z minus 1 into z plus 1 whole square.
01:33
These two will get cancelled out.
01:35
Now putting the limit this will be 1 over 1 plus 1 whole square which is 2 square.
01:41
So that means residue at z equal to 1 will be 1 by 4.
01:48
Now let us find out residue at z equal to minus 1...