Final answer:
To represent f(z):=1 / ((z+1)(z-3)) by a Laurent series on the given annuli, split f(z) into partial fractions, calculate the coefficients, and expand each term as a power series about its singularity. The final Laurent series is the sum of these series, converging in the annulus 1 < |z-2| < 3.
Step-by-step explanation:
The student asked to represent the function f(z):=1 / ((z+1)(z-3)) by a Laurent series on the annuli A:=z-2. To find this Laurent series, we need to decompose the function into simpler parts that we can expand. Notice that f(z) has singularities at z=-1 and z=3, so we aim to split f(z) into partial fractions that can be more easily handled.
First, we perform partial fraction decomposition:
f(z) = 1/((z+1)(z-3)) = A/(z+1) + B/(z-3)
By solving the equations for coefficients A and B, we get A = -1/4 and B = 1/4. Now we write:
f(z) = -1/4(z+1) + 1/4(z-3)
Next, express each fraction as a power series by geometric series expansion within the annuli A:
-1/4(z+1) can be expanded as a Laurent series about z=-1, and 1/4(z-3) can be expanded about z=3. Summing the resulting series gives us the Laurent series representation of f(z) on the annuli A.
The series obtained will only converge in the annulus 1 < |z-2| < 3 because of the positions of the singularities and the region mentioned.