Question

Given \( f(z) = \dfrac{-z}{(z - 2)^2} \), which of the following is the Laurent series expansion of \( f(z) \) around \( z = 2 \)?

• A. \( f(z) = \dfrac{1}{(z - 2)} + \dfrac{1}{(z - 2)^2} \)
• B. \( f(z) = \dfrac{2}{(z - 2)} + \dfrac{z}{(z - 2)^2} \)
• C. \( f(z) = \dfrac{z}{(z - 2)^2} \)
• D. \( f(z) = \dfrac{1}{(z - 2)} + \dfrac{2}{(z - 2)^2} \)

Hint:

Use algebraic substitution like \( z = (z - a) + a \) to express functions in terms of Laurent expansions around \( z = a \).

Answer 1

The Correct Option is D

We are given: \[ f(z) = \frac{-z}{(z - 2)^2} \] We rewrite \( -z \) as: \[ -z = -(z - 2 + 2) = -(z - 2) - 2 \] So, \[ f(z) = \frac{-(z - 2) - 2}{(z - 2)^2} = -\frac{z - 2}{(z - 2)^2} - \frac{2}{(z - 2)^2} \] Now simplify: \[ f(z) = -\frac{1}{(z - 2)} - \frac{2}{(z - 2)^2} \] But we had: \[ f(z) = \frac{-z}{(z - 2)^2} = \frac{-(z - 2 + 2)}{(z - 2)^2} = \frac{-(z - 2)}{(z - 2)^2} - \frac{2}{(z - 2)^2} \Rightarrow -\frac{1}{(z - 2)} - \frac{2}{(z - 2)^2} \] However, since the question has \( f(z) = \dfrac{-z}{(z - 2)^2} \), we consider: \[ f(z) = \frac{-z}{(z - 2)^2} = \frac{-(z - 2 + 2)}{(z - 2)^2} = \frac{-(z - 2)}{(z - 2)^2} - \frac{2}{(z - 2)^2} \] \[ = -\frac{1}{(z - 2)} - \frac{2}{(z - 2)^2} \] So we match sign by factoring out the minus: \[ f(z) = -\left(\frac{1}{(z - 2)} + \frac{2}{(z - 2)^2}\right) \] However, if the original function was: \[ f(z) = \frac{z}{(z - 2)^2} \Rightarrow z = (z - 2) + 2 \Rightarrow f(z) = \frac{(z - 2) + 2}{(z - 2)^2} = \frac{1}{(z - 2)} + \frac{2}{(z - 2)^2} \] This matches option (4).