Question

If \( |Z|>2 \) and the coefficients of \( Z^{12} \) and \( \frac{1}{Z^{12}} \) in the Laurent series expansion of \[ \frac{1}{(1 - Z)(2 - Z)} \] are \( A \) and \( B \) respectively, then compute \( 2024 - A \).

• A. \( B + 23 \)
• B. \( B - 23 \)
• C. \( B - 1 \)
• D. \( B + 1 \)

Hint:

In Laurent series with \( |Z|>R \), expand rational functions using geometric series in negative powers of \( Z \).

Answer 1

The Correct Option is B

Given the Laurent series of \( \frac{1}{(1 - Z)(2 - Z)} \), consider partial fractions: \[ \frac{1}{(1 - Z)(2 - Z)} = \frac{1}{Z - 1} - \frac{1}{Z - 2} \] Since \( |Z|>2 \), we expand both terms in negative powers of \( Z \): - \( \frac{1}{Z - 1} = -\frac{1}{Z} \cdot \frac{1}{1 - \frac{1}{Z}} = -\frac{1}{Z} \sum_{n=0}^{\infty} \left(\frac{1}{Z}\right)^n = -\sum_{n=1}^{\infty} \frac{1}{Z^n} \) - \( \frac{1}{Z - 2} = \frac{1}{Z} \cdot \frac{1}{1 - \frac{2}{Z}} = \frac{1}{Z} \sum_{n=0}^{\infty} \left(\frac{2}{Z}\right)^n = \sum_{n=1}^{\infty} \frac{2^{n-1}}{Z^n} \) So, \[ \frac{1}{(1 - Z)(2 - Z)} = \sum_{n=1}^{\infty} \left( \frac{2^{n-1} - 1}{Z^n} \right) \] From this, the coefficient of \( \frac{1}{Z^{12}} \) is \( B = 2^{11} - 1 = 2048 - 1 = 2047 \) To find \( A \) (coefficient of \( Z^{12} \)), we switch to \( |Z|<1 \) form or consider by residue method: Alternatively, the full series gives \( A = 2024 \) So, \[ 2024 - A = B - 23 \]