Question

Consider the function \( f(z) = \frac{2z + 1}{z^2 - 2z} \), where \( z \) is a complex variable. The sum of the residues at singular points of \( f(z) \) is:

Hint:

The sum of residues of a function inside a closed contour is equal to the sum of residues at the singularities inside that contour. This is a result of the residue theorem.

Answer 1

Step 1: Identify singular points. 
The function has singularities where the denominator is zero: \[ z^2 - 2z = 0 \quad \Rightarrow \quad z(z - 2) = 0 \] So, the singular points are at \( z = 0 \) and \( z = 2 \). 
Step 2: Calculate residues at each singular point. For \( z = 0 \), we can write the function as: \[ f(z) = \frac{2z + 1}{z(z - 2)} \] To find the residue at \( z = 0 \), we evaluate the limit: \[ \text{Residue at } z = 0 = \lim_{z \to 0} z \cdot f(z) = \lim_{z \to 0} \frac{2z + 1}{z - 2} = \frac{1}{-2} = -\frac{1}{2} \] For \( z = 2 \), we can rewrite the function as: \[ f(z) = \frac{2z + 1}{(z - 2)z} \] To find the residue at \( z = 2 \), we evaluate the limit: \[ \text{Residue at } z = 2 = \lim_{z \to 2} (z - 2) \cdot f(z) = \lim_{z \to 2} \frac{2z + 1}{z} = \frac{2(2) + 1}{2} = \frac{5}{2} \] Step 3: Sum of residues. 
The sum of the residues at the singular points is: \[ \text{Sum of residues} = -\frac{1}{2} + \frac{5}{2} = \frac{4}{2} = 2 \]