Question

The Laurent series of $ f(z) = \frac{z}{(z^2+1)(z^2+4)} $ for $ |z|<1 $ is:

• A. \( \frac{1}{4} z - \frac{5}{16} z^3 + \frac{21}{64} z^5 + \cdots \)
• B. \( \frac{1}{2} - \frac{1}{4} z^2 + \frac{5}{16} z^4 + \frac{21}{64} z^6 + \cdots \)
• C. \( \frac{1}{2} z - \frac{3}{4} z^3 + \frac{15}{8} z^5 + \cdots \)
• D. \( \frac{1}{2} + \frac{1}{2} z^2 + \frac{3}{4} z^4 + \frac{15}{8} z^6 + \cdots \)

Hint:

Double-checking the partial fraction decomposition is crucial. Ensure the coefficients are calculated correctly before expanding into geometric series.

Answer 1

The Correct Option is A

Step 1: Perform partial fraction decomposition of \( f(z) \). 
We have \( f(z) = \frac{z}{(z^2+1)(z^2+4)} \). Let \( w = z^2 \). Then \( \frac{w}{(w+1)(w+4)} = \frac{A}{w+1} + \frac{B}{w+4} \).
Multiplying by \( (w+1)(w+4) \), we get \( w = A(w+4) + B(w+1) \).
Setting \( w = -1 \), we get \( -1 = 3A \implies A = -\frac{1}{3} \).
Setting \( w = -4 \), we get \( -4 = -3B \implies B = \frac{4}{3} \).
So, \( \frac{w}{(w+1)(w+4)} = -\frac{1}{3} \frac{1}{w+1} + \frac{4}{3} \frac{1}{w+4} \).
Substituting \( w = z^2 \), we get \( f(z) = -\frac{1}{3} \frac{1}{z^2+1} + \frac{4}{3} \frac{1}{z^2+4} \). 
Step 2: Expand each term as a geometric series since \( |z|<1 \). 
For the first term: \[ -\frac{1}{3} \frac{1}{z^2+1} = -\frac{1}{3} \frac{1}{1 - (-z^2)} = -\frac{1}{3} \sum_{n=0}^{\infty} (-z^2)^n = -\frac{1}{3} \sum_{n=0}^{\infty} (-1)^n z^{2n} = -\frac{1}{3} (1 - z^2 + z^4 - z^6 + \cdots) \] For the second term: \[ \frac{4}{3} \frac{1}{z^2+4} = \frac{4}{3} \frac{1}{4(1 + \frac{z^2}{4})} = \frac{1}{3} \frac{1}{1 - (-\frac{z^2}{4})} = \frac{1}{3} \sum_{n=0}^{\infty} (-\frac{z^2}{4})^n = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{4^n} \] \[ = \frac{1}{3} (1 - \frac{z^2}{4} + \frac{z^4}{16} - \frac{z^6}{64} + \cdots) \] 
Step 3: Combine the two series. 
\[ f(z) = -\frac{1}{3} (1 - z^2 + z^4 - z^6 + \cdots) + \frac{1}{3} (1 - \frac{z^2}{4} + \frac{z^4}{16} - \frac{z^6}{64} + \cdots) \] \[ f(z) = (-\frac{1}{3} + \frac{1}{3}) + (\frac{1}{3} - \frac{1}{12}) z^2 + (-\frac{1}{3} + \frac{1}{48}) z^4 + (\frac{1}{3} - \frac{1}{192}) z^6 + \cdots \] \[ f(z) = 0 + (\frac{4-1}{12}) z^2 + (\frac{-16+1}{48}) z^4 + (\frac{64-1}{192}) z^6 + \cdots \] \[ f(z) = \frac{3}{12} z^2 - \frac{15}{48} z^4 + \frac{63}{192} z^6 + \cdots = \frac{1}{4} z^2 - \frac{5}{16} z^4 + \frac{21}{64} z^6 + \cdots \] This still seems incorrect based on the provided answer. Let's re-do the partial fraction for \( f(z)/z \). \( \frac{1}{(z^2+1)(z^2+4)} = \frac{A}{z^2+1} + \frac{B}{z^2+4} \) \( 1 = A(z^2+4) + B(z^2+1) \) If \( z^2 = -1 \), \( 1 = A(3) \implies A = 1/3 \).
If \( z^2 = -4 \), \( 1 = B(-3) \implies B = -1/3 \).
So, \( \frac{f(z)}{z} = \frac{1}{3} \left( \frac{1}{z^2+1} - \frac{1}{z^2+4} \right) \) \( f(z) = \frac{z}{3} \left( \frac{1}{1-(-z^2)} - \frac{1}{4(1-(-z^2/4))} \right) \) \( f(z) = \frac{z}{3} \left( \sum_{n=0}^\infty (-z^2)^n - \frac{1}{4} \sum_{n=0}^\infty (-\frac{z^2}{4})^n \right) \) 
Coefficient of \( z^1 \) (\( n=0 \) term): \( \frac{1}{3} (1) - \frac{1}{12} (1) = \frac{4-1}{12} = \frac{3}{12} = \frac{1}{4} \).
Coefficient of \( z^3 \) (\( n=1 \) term): \( \frac{1}{3} (-1) - \frac{1}{12} (-\frac{1}{4}) = -\frac{1}{3} + \frac{1}{48} = \frac{-16+1}{48} = -\frac{15}{48} = -\frac{5}{16} \).
Coefficient of \( z^5 \) (\( n=2 \) term): \( \frac{1}{3} (1) - \frac{1}{12} (\frac{1}{16}) = \frac{1}{3} - \frac{1}{192} = \frac{64-1}{192} = \frac{63}{192} = \frac{21}{64} \).
So, \( f(z) = \frac{1}{4} z - \frac{5}{16} z^3 + \frac{21}{64} z^5 + \cdots \)