Final answer:
The task is to write f(z) as partial fractions and integrate over the unit circle. Since the denominator of f(z) = (z + 1)/(z² + 2) is irreducible, partial fractions are not needed, and the integral around the unit circle is zero by Cauchy's integral theorem.
Step-by-step explanation:
The objective is to write the function f(z) = ··· in terms of partial fractions and then integrate it counterclockwise over the unit circle. However, there may be some confusion because the function provided, f(z) = z + 1/z² + 2, needs clarification. If we assume it's f(z) = (z + 1)/(z² + 2), then we can proceed with the partial fraction decomposition and complex integration.
First, we perform partial fractions decomposition. However, in this case, the denominator is already irreducible since it's a quadratic with no real roots, which means we don't need to decompose it further for the integration over the unit circle.
For the integral over the unit circle, one might use the residue theorem from complex analysis, which states that the integral of a function around a closed curve is 2πi times the sum of the residues of the function within that curve. However, since the only pole of f(z) within the unit circle would be at z = 0, which is not a pole of the function as given, the integral is zero due to Cauchy's integral theorem.