Question

For the complex function f(z) = (z² - 6z + 5)/(z³(z + 3)), (a) determine the poles and their order. (b) calculate the residue at each pole.

Answer 1

(a) The function f(z) has poles at z = 0 of order 3 and z = 3 of order 1.

(b) The residues at the poles are Res(z=0) = -1/9 and Res(z=3) = 2/27.

Step-by-step explanation:

The function f(z) can be expressed as f(z) = (z² - 6z + 5)/(z³(z + 3)). To determine the poles and their orders, we need to find the values of z for which the denominator becomes zero. Setting the denominator equal to zero gives us z = 0 and z = -3 as the poles. Since z = 0 occurs three times in the denominator, it is a pole of order 3. On the other hand, z = -3 occurs once in the denominator, making it a pole of order 1.

To calculate the residues at each pole, we can use the formula Res(z=a) = lim(z→a) [(z-a)^n f(z)], where n is the order of the pole at z=a. For z=0, the residue is Res(z=0) = lim(z→0) [((z-0)^3 (z² - 6z + 5))/(z³(z + 3))]. After simplifying, we get Res(z=0) = -1/9. Similarly, for z=3, the residue is Res(z=3) = lim(z→3) [(z-3)*(z² - 6z + 5)/(z³(z + 3))], which simplifies to Res(z=3) = 2/27.

In summary, the function f(z) has poles at z = 0 of order 3 and z = 3 of order 1. The residues at these poles are Res(z=0) = -1/9 and Res(z=3) = 2/27.

Answer 2

The poles of the function f(z) are at z = 0 (order 3) and z = -3 (order 1). Residues at these poles can be computed by differentiating for the higher-order pole and taking limits for the pole of order 1.

Step-by-step explanation:

The complex function given is f(z) = (z² - 6z + 5)/(z³(z + 3)). To find the poles and their order, we look for values of z that make the denominator zero, as those are the points where the function goes to infinity. The denominator factors as z³(z + 3), so the poles are at z = 0 and z = -3. The pole at z = 0 is of order 3 because the factor z³ in the denominator has a multiplicity of 3. The pole at z = -3 is of order 1 because the factor (z + 3) has a multiplicity of 1.

To calculate the residue at each pole, we apply different methods depending on the order of the pole. For the pole of order 1 at z = -3, the residue is the coefficient of 1/(z + 3) in the Laurent series expansion. To find the residue at the pole of order 3 at z = 0, we use the formula for the residue of higher-order poles which involves taking derivatives. The calculations are:

• Residue at z = -3: Simplify f(z) by multiplying and dividing by (z + 3) and calculate the limit as z approaches -3.
• Residue at z = 0: Calculate the coefficient of 1/z in the Laurent series by differentiating the reduced function f(z) twice and evaluating at z = 0.