Expand the following functions into Laurent series in the indicated regions:
\(\displaystyle f(z) = \frac{z^2 - 1}{ (z+2)(z+3)}, \; \; 2 < |z| < 3\), \(3 < |z| < + \infty\).
\(\displaystyle f(z) = \sin \frac{z}{1-z}, \; \; 0 < |z-1| < + \infty\)
Solution
Goal: Compute the exact Laurent series expansions for each function in the specified annular domains.
1. Part (a): Partial fraction decomposition of \(f(z)\).Proof:
1.1. Perform polynomial division on \(f(z) = \frac{z^2 - 1}{z^2 + 5z + 6}\):\[\frac{z^2 - 1}{z^2 + 5z + 6} = 1 + \frac{-5z - 7}{(z+2)(z+3)}.\]
1.2. Decompose the remainder \(\frac{-5z - 7}{(z+2)(z+3)} = \frac{A}{z+2} + \frac{B}{z+3}\):\[A = \lim_{z \to -2} (z+2) \frac{-5z - 7}{(z+2)(z+3)} = \frac{10 - 7}{1} = 3,\]\[B = \lim_{z \to -3} (z+3) \frac{-5z - 7}{(z+2)(z+3)} = \frac{15 - 7}{-1} = -8.\]
1.3. Thus the partial fraction expansion is\[f(z) = 1 + \frac{3}{z+2} - \frac{8}{z+3}.\]
2. Part (a): Laurent series in the annulus \(2 < |z| < 3\).Proof:
2.1. For \(|z| > 2\), \(|2/z| < 1\), so we expand \(\frac{3}{z+2}\) in powers of \(1/z\):\[\frac{3}{z+2} = \frac{3}{z} \frac{1}{1 + 2/z} = \frac{3}{z} \sum_{k=0}^\infty \left(-\frac{2}{z}\right)^k = 3 \sum_{k=0}^\infty (-2)^k z^{-k-1} = \sum_{j=1}^\infty 3(-2)^{j-1} z^{-j}.\]
2.2. For \(|z| < 3\), \(|z/3| < 1\), so we expand \(-\frac{8}{z+3}\) in positive powers of \(z\):\[-\frac{8}{z+3} = -\frac{8}{3} \frac{1}{1 + z/3} = -\frac{8}{3} \sum_{k=0}^\infty \left(-\frac{z}{3}\right)^k = -8 \sum_{k=0}^\infty \frac{(-1)^k}{3^{k+1}} z^k.\]
2.3. Combining the constant term and both series:\[f(z) = 1 - \frac{8}{3} - 8 \sum_{k=1}^\infty \frac{(-1)^k}{3^{k+1}} z^k + 3 \sum_{j=1}^\infty (-2)^{j-1} z^{-j} = -\frac{5}{3} - 8 \sum_{k=1}^\infty \frac{(-1)^k}{3^{k+1}} z^k + \sum_{j=1}^\infty 3(-2)^{j-1} z^{-j}.\]
3. Part (a): Laurent series in the exterior domain \(3 < |z| < \infty\).Proof:
3.1. Since \(|z| > 3 > 2\), both \(|2/z| < 1\) and \(|3/z| < 1\), so both fractions are expanded in negative powers of \(z\):\[\frac{3}{z+2} = 3 \sum_{k=0}^\infty (-2)^k z^{-k-1} = \sum_{j=1}^\infty 3(-2)^{j-1} z^{-j},\]\[-\frac{8}{z+3} = -\frac{8}{z} \frac{1}{1 + 3/z} = -8 \sum_{k=0}^\infty (-3)^k z^{-k-1} = -\sum_{j=1}^\infty 8(-3)^{j-1} z^{-j}.\]
3.2. Combining these with the constant 1 gives\[f(z) = 1 + \sum_{j=1}^\infty \left[ 3(-2)^{j-1} - 8(-3)^{j-1} \right] z^{-j}.\]
4. Part (b): Laurent series for \(\sin\left(\frac{z}{1-z}\right)\) in \(0 < |z-1| < \infty\).Proof:
4.1. Let \(w = z - 1\), so \(z = 1 + w\) and \(1 - z = -w\). The region becomes the punctured plane \(0 < |w| < \infty\).
4.2. The argument simplifies to\[\frac{z}{1-z} = \frac{1 + w}{-w} = -1 - \frac{1}{w}.\]
4.3. Using the angle addition formula for sine:\[\sin\left(-1 - \frac{1}{w}\right) = -\sin\left(1 + \frac{1}{w}\right) = -\sin(1)\cos\left(\frac{1}{w}\right) - \cos(1)\sin\left(\frac{1}{w}\right).\]
4.4. Substitute the entire Taylor series for cosine and sine evaluated at \(1/w\):\[\cos\left(\frac{1}{w}\right) = \sum_{m=0}^\infty \frac{(-1)^m}{(2m)!} w^{-2m}, \qquad \sin\left(\frac{1}{w}\right) = \sum_{m=0}^\infty \frac{(-1)^m}{(2m+1)!} w^{-(2m+1)}.\]
4.5. Substituting \(w = z - 1\) gives the Laurent series valid for all \(0 < |z-1| < \infty\):\[\sin\left(\frac{z}{1-z}\right) = -\sin(1) \sum_{m=0}^\infty \frac{(-1)^m}{(2m)!} (z-1)^{-2m} - \cos(1) \sum_{m=0}^\infty \frac{(-1)^m}{(2m+1)!} (z-1)^{-(2m+1)}.\]
5. Conclusion:Proof:All requested Laurent expansions are completely determined.