Let's solve each part **step-by-step**: --- ### (7) Let \( f(z) = \frac{1}{(2z+1)(z+3)} \). #### (a) **Maclaurin Series Representation (about \( z = 0 \)); Region of Convergence** **Step 1: Partial Fractions** \[ f(z) = \frac{1}{(2z+1)(z+3)} = \frac{A}{2z+1} + \frac{B}{z+3} \] Multiply both sides by \((2z+1)(z+3)\): \[ 1 = A(z+3) + B(2z+1) \] Let’s solve for \(A\) and \(B\): Expand: \[ 1 = A(z+3) + B(2z+1) = Az + 3A + 2Bz + B = (A + 2B)z + (3A + B) \] Equate coefficients: - \(z\) terms: \(A + 2B = 0\) - Constant: \(3A + B = 1\) From \(A + 2B = 0\): \(A = -2B\) Substitute into the second equation: \[ 3(-2B) + B = 1 \implies -6B + B = 1 \implies -5B = 1 \implies B = -\frac{1}{5} \] \[ A = -2B = -2 \left(-\frac{1}{5}\right) = \frac{2}{5} \] So, \[ f(z) = \frac{2}{5} \frac{1}{2z+1} - \frac{1}{5} \frac{1}{z+3} \] --- **Step 2: Maclaurin Series for Each Term** Write in the form \(\frac{1}{1-w}\): - \(\frac{1}{2z+1} = \frac{1}{1+2z}\) - \(\frac{1}{z+3} = \frac{1}{3(1+z/3)}\) So, \[ f(z) = \frac{2}{5} \cdot \frac{1}{1+2z} - \frac{1}{5} \cdot \frac{1}{3} \cdot \frac{1}{1+z/3} = \frac{2}{5} \sum_{n=0}^\infty (-2z)^n - \frac{1}{15} \sum_{n=0}^\infty (-z/3)^n \] \[ = \frac{2}{5} \sum_{n=0}^{\infty} (-2)^n z^n - \frac{1}{15} \sum_{n=0}^{\infty} \left(-\frac{1}{3}\right)^n z^n \] \[ = \sum_{n=0}^{\infty} \left[\frac{2}{5}(-2)^n - \frac{1}{15}\left(-\frac{1}{3}\right)^n \right] z^n \] --- **Region of Convergence:** - \(\frac{1}{1+2z}\) converges for \(|2z| < 1 \implies |z| < \frac{1}{2}\) - \(\frac{1}{1+z/3}\) converges for \(|z/3| < 1 \implies |z| < 3\) So, common region: \(|z| < \frac{1}{2}\) --- **Final Answer (a):** \[ \boxed{ f(z) = \sum_{n=0}^{\infty} \left[\frac{2}{5}(-2)^n - \frac{1}{15}\left(-\frac{1}{3}\right)^n \right] z^n, \quad \text{convergent for } |z| < \frac{1}{2} } \] --- #### (b) **Laurent Series Expansion in \( 0 < |z+\frac{1}{2}| < \frac{5}{2} \)** Here, expand about \( z = -\frac{1}{2} \). Let \( w = z + \frac{1}{2} \implies z = w - \frac{1}{2} \): Substitute into \( f(z) \): \[ 2z+1 = 2(w-\frac{1}{2})+1 = 2w-1+1 = 2w \] \[ z+3 = w - \frac{1}{2} + 3 = w + \frac{5}{2} \] So, \[ f(z) = \frac{1}{(2w)(w+\frac{5}{2})} = \frac{1}{2w(w+\frac{5}{2})} \] Let's use partial fractions in \(w\): \[ \frac{1}{2w(w+\frac{5}{2})} = \frac{A}{w} + \frac{B}{w+\frac{5}{2}} \] Multiply both sides by \(2w(w+\frac{5}{2})\): \[ 1 = 2A(w+\frac{5}{2}) + 2B w \] Let’s solve for \(A\) and \(B\): Expand: \[ 1 = 2A w + 5A + 2B w = (2A + 2B)w + 5A \] Equate coefficients: - \(w\) terms: \(2A + 2B = 0\) \(\implies A = -B\) - Constant: \(5A = 1\) \(\implies A = \frac{1}{5}, B = -\frac{1}{5}\) So, \[ \frac{1}{2w(w+\frac{5}{2})} = \frac{1}{5w} - \frac{1}{5(w+\frac{5}{2})} \] Now, expand \( \frac{1}{w+\frac{5}{2}} \) in powers of \( \frac{w}{5/2} \): \[ \frac{1}{w+\frac{5}{2}} = \frac{1}{\frac{5}{2}} \cdot \frac{1}{1 + \frac{w}{5/2}} = \frac{2}{5} \sum_{n=0}^{\infty} (-1)^n \left( \frac{w}{5/2} \right)^n, \quad \left|\frac{w}{5/2}\right| < 1 \] \[ = \frac{2}{5} \sum_{n=0}^{\infty} (-1)^n \left( \frac{2w}{5} \right)^n \] So, \[ f(z) = \frac{1}{5w} - \frac{1}{5} \cdot \frac{2}{5} \sum_{n=0}^{\infty} (-1)^n \left( \frac{2w}{5}\right)^n = \frac{1}{5w} - \frac{2}{25} \sum_{n=0}^{\infty} (-1)^n \left( \frac{2w}{5} \right)^n \] \[ = \frac{1}{5w} - \frac{2}{25} \sum_{n=0}^{\infty} (-1)^n \left( \frac{2}{5} \right)^n w^n \] This is the Laurent series for \(0 < |w| < \frac{5}{2}\), i.e., \(0 < |z + \frac{1}{2}| < \frac{5}{2}\). --- **Final Answer (b):** \[ \boxed{ f(z) = \frac{1}{5(z+\frac{1}{2})} - \frac{2}{25} \sum_{n=0}^{\infty} \left[ (-1)^n \left( \frac{2}{5} \right)^n (z+\frac{1}{2})^n \right],\quad 0 < |z+\frac{1}{2}| < \frac{5}{2} } \] --- #### (c) **Laurent Series in Powers of \( z \) Convergent at \( z = 5 \); Region of Convergence** The singularities are at \( z = -\frac{1}{2} \) and \( z = -3 \). Distance from \( z = 5 \) to these points: - To \( z = -\frac{1}{2} \): \( |5 + \frac{1}{2}| = 5.5 \) - To \( z = -3 \): \( |5 + 3| = 8 \) The series in powers of \( 1/z \) (Laurent expansion about infinity) will converge for \( |z| > 3 \) (outside the outermost singularity). Let’s write \( f(z) \) as a Laurent series about \( z = \infty \): From earlier: \[ f(z) = \frac{1}{(2z+1)(z+3)} = \frac{2}{5} \frac{1}{2z+1} - \frac{1}{5} \frac{1}{z+3} \] Expand each for large \(z\): - \( \frac{1}{2z+1} = \frac{1}{2z} \cdot \frac{1}{1 + \frac{1}{2z}} = \frac{1}{2z} \sum_{n=0}^\infty (-1)^n \left( \frac{1}{2z} \right)^n \) - \( \frac{1}{z+3} = \frac{1}{z} \cdot \frac{1}{1 + \frac{3}{z}} = \frac{1}{z} \sum_{n=0}^\infty (-1)^n \left( \frac{3}{z} \right)^n \) So, \[ f(z) = \frac{2}{5} \cdot \frac{1}{2z} \sum_{n=0}^\infty (-1)^n \left( \frac{1}{2z} \right)^n - \frac{1}{5} \cdot \frac{1}{z} \sum_{n=0}^\infty (-1)^n \left( \frac{3}{z} \right)^n \] Simplify: First term: \[ \frac{2}{5} \cdot \frac{1}{2z} \sum_{n=0}^\infty (-1)^n \left( \frac{1}{2z} \right)^n = \frac{1}{5z} \sum_{n=0}^\infty (-1)^n \left( \frac{1}{2z} \right)^n = \frac{1}{5z} \sum_{n=0}^\infty (-1)^n (2^{-n}) z^{-n} \] Second term: \[ - \frac{1}{5} \cdot \frac{1}{z} \sum_{n=0}^\infty (-1)^n \left( \frac{3}{z} \right)^n = -\frac{1}{5z} \sum_{n=0}^\infty (-1)^n 3^n z^{-n} \] Combine: \[ f(z) = \frac{1}{5z} \sum_{n=0}^\infty (-1)^n \left[ 2^{-n} - 3^n \right] z^{-n} \] Or, reindex: \[ f(z) = \sum_{n=0}^\infty \frac{(-1)^n}{5} \left[ 2^{-n} - 3^n \right] z^{-n-1} \] --- **Region of convergence:** \( |z| > 3 \) (outside the outermost singularity). --- **Final Answer (c):** \[ \boxed{ f(z) = \sum_{n=0}^\infty \frac{(-1)^n}{5} \left[2^{-n} - 3^n\right] z^{-n-1}, \quad |z| > 3 } \] --- **Summary Table:** | Part | Series | Region of Convergence | |------|--------|----------------------| | (a) | \( \sum_{n=0}^{\infty} \left[\frac{2}{5}(-2)^n - \frac{1}{15}(-\frac{1}{3})^n\right]z^n \) | \( |z| < \frac{1}{2} \) | | (b) | \( \frac{1}{5(z+\frac{1}{2})} - \frac{2}{25} \sum_{n=0}^{\infty} (-1)^n (\frac{2}{5})^n (z+\frac{1}{2})^n \) | \( 0 < |z+\frac{1}{2}| < \frac{5}{2} \) | | (c) | \( \sum_{n=0}^\

(a) \(\boxed{f(z) = \sum_{n=0}^{\infty} \left[\frac{2}{5}(-2)^n - \frac{1}{15}\left(-\frac{1}{3}\right)^n \right] z^n,\quad |z|<\frac{1}{2}}\) (b) \(\boxed{f(z) = \frac{1}{5(z+\frac{1}{2})} - \frac{2}{25} \sum_{n=0}^{\infty} (-1)^n \left(\frac{2}{5}\right)^n (z+\frac{1}{2})^n,\quad 0<|z+\frac{1}{2}|<\frac{5}{2}}\) (c) \(\boxed{f(z) = \sum_{n=0}^\infty \frac{(-1)^n}{5} \left(2^{-n} - 3^n\right) z^{-n-1},\quad |z|>3}\)