Question

Consider the function \( f(z) = \frac{1}{z^2 (z - 2)^3} \) of a complex variable \( z \).
The residues of the function at \( z = 0 \) and \( z = 2 \), respectively, are:

• A. \( -\frac{3}{8} \) and \( \frac{3}{8} \)
• B. \( \frac{3}{8} \) and \( -\frac{3}{16} \)
• C. \( -\frac{3}{16} \) and \( \frac{3}{16} \)
• D. \( -\frac{3}{8} \) and \( \frac{3}{16} \)

Hint:

For finding residues, remember to use the appropriate formula based on the order of the pole (second-order or third-order) and apply differentiation if needed.

Answer 1

The Correct Option is C

The function \( f(z) = \frac{1}{z^2 (z - 2)^3} \) has singularities at \( z = 0 \) and \( z = 2 \). We need to compute the residues of this function at both points. 
Step 1: Residue at \( z = 0 \): 
At \( z = 0 \), we have a pole of order 2. To find the residue, we use the formula for a second-order pole: \[ {Res}(f, 0) = \lim_{z \to 0} \frac{d}{dz} \left[ z^2 f(z) \right]. \] Now, substitute \( f(z) = \frac{1}{z^2 (z - 2)^3} \): \[ z^2 f(z) = \frac{1}{(z - 2)^3}. \] Differentiating with respect to \( z \): \[ \frac{d}{dz} \left[ \frac{1}{(z - 2)^3} \right] = -3 \cdot \frac{1}{(z - 2)^4}. \] At \( z = 0 \), this becomes: \[ {Res}(f, 0) = \frac{-3}{16}. \] Step 2: Residue at \( z = 2 \): 
At \( z = 2 \), we have a pole of order 3. To compute the residue, we use the formula for the residue of a third-order pole: \[ {Res}(f, 2) = \lim_{z \to 2} \frac{1}{2!} \frac{d^2}{dz^2} \left[ (z - 2)^3 f(z) \right]. \] Substitute \( f(z) = \frac{1}{z^2 (z - 2)^3} \): \[ (z - 2)^3 f(z) = \frac{1}{z^2}. \] Differentiating twice with respect to \( z \): \[ \frac{d^2}{dz^2} \left[ \frac{1}{z^2} \right] = \frac{6}{z^4}. \] At \( z = 2 \), this becomes: \[ {Res}(f, 2) = \frac{6}{16} = \frac{3}{16}. \] Thus, the residues are \( {Res}(f, 0) = -\frac{3}{16} \) and \( {Res}(f, 2) = \frac{3}{16} \).