Question

Consider the function \( f(z) = \frac{1}{(z+1)(z+2)(z+3)} \). The residue of \( f(z) \) at \( z = -1 \) is _________.

Hint:

To calculate the residue at a simple pole, multiply the function by \( (z - z_0) \) and evaluate the limit as \( z \to z_0 \).

Answer 1

Correct Answer: 0.49

We are given the function: \[ f(z) = \frac{1}{(z+1)(z+2)(z+3)}. \] Step 1: Identify the singularity
The singularities of the function are at \( z = -1 \), \( z = -2 \), and \( z = -3 \). We are interested in the residue at \( z = -1 \). Step 2: Calculate the residue
The residue at \( z = -1 \) is given by: \[ \text{Res}(f, -1) = \lim_{z \to -1} (z + 1) f(z). \] Substitute \( f(z) \) into the equation: \[ \text{Res}(f, -1) = \lim_{z \to -1} \frac{1}{(z+2)(z+3)}. \] Now, substitute \( z = -1 \): \[ \text{Res}(f, -1) = \frac{1}{(-1+2)(-1+3)} = \frac{1}{(1)(2)} = \frac{1}{2}. \] Thus, the residue of \( f(z) \) at \( z = -1 \) is \( 0.5 \).