Final answer:
The residues of f(z)=3z/(z+2)²(z²−1) are 1/3 at z=1, -3 at z=-1, and 0 at z=-2. You calculate these by using the limit of (z - pole) multiplied by f(z) for simple poles, and for a double pole by taking the derivative of the adjusted function and evaluating the limit.
Step-by-step explanation:
The question asks to find the residues at all points of f(z)=3z/(z+2)²(z²−1). To find the residues, we need to identify the poles of the function and then calculate the residue at each pole. The poles of the function are the values of z where the denominator is zero, which are z=-2 (a pole of order 2), z=1, and z=-1 (both simple poles).
The residue at a simple pole can be found by taking the limit as z approaches the pole of the function multiplied by (z - pole). For the double pole at z=-2, we find the residue by taking the derivative of the function after multiplying by (z+2)2 and then taking the limit as z approaches -2. Let's calculate:
Residue at z=1: limz→1 (z-1)f(z) = 3z/(z+2)2 = 3/9 = 1/3
Residue at z=-1: limz→-1 (z+1)f(z) = -3z/(z+2)2 = -3/1 = -3
Residue at z=-2: Need to compute the derivative of 3z as the other factor is a constant at z=-2, which gives us a residue of 0
Therefore, the residues at z=1, z=-1, and z=-2 are 1/3, -3, and 0, respectively.