Question

Let \( D = \{ z \in \mathbb{C} : |z| < 2 \pi \} \) and \( f: D \to \mathbb{C} \) be the function defined by \[ f(z) = \begin{cases} \frac{3z^2}{1 - \cos z} & \text{if } z \neq 0, \\ 6 & \text{if } z = 0. \end{cases} \] If \( f(z) = \sum_{n=0}^{\infty} a_n z^n \text{ for } z \in D, \text{ then } 6a_2 = \underline{\hspace{1cm}}. \)

Hint:

To find the coefficients in the power series, expand the function using known series expansions and match terms.

Answer 1

Correct Answer: 3

We are tasked with finding the coefficient \( a_2 \) in the power series expansion of \( f(z) \) around \( z = 0 \). The function is given piecewise, so we begin by expanding the function for \( z \neq 0 \) and compute the power series of \( \frac{3z^2}{1 - \cos z} \). Using the Maclaurin series expansion of \( \cos z \): \[ \cos z = 1 - \frac{z^2}{2} + \frac{z^4}{24} - \cdots \] we get: \[ 1 - \cos z = \frac{z^2}{2} - \frac{z^4}{24} + \cdots \] Thus, the series for \( f(z) \) becomes: \[ f(z) = \frac{3z^2}{\frac{z^2}{2} - \frac{z^4}{24} + \cdots} = 6 + \cdots \] By matching terms, we find that: \[ 6a_2 = 3. \] Thus, \( 6a_2 = \boxed{3} \).