Find the series of the following function in the given domain.
- $f(z) = \frac{z^2+2z-4}{(z^2-9)(z+1)}, 1 < |z| < 3$.
- $f(z) = \frac{1}{z^2+1}, 0 < |z+i| < 2$.
Attempt:
Notice that \begin{equation*} \frac{z^2+2z-4}{(z^2-9)(z+1)} = \frac{5}{8} \cdot \frac{1}{z+1} - \frac{1}{12} \cdot \frac{1}{z+3} + \frac{11}{24} \cdot \frac{1}{z-3}. \end{equation*} For $1 < |z|$, we have $\frac{1}{|z|} < 1$. For $|z| < 3$, we have $\left|\frac{z}{3} \right| < 1$. Hence, for $1 < |z| < 3$, we have \begin{align*} f(z) &= \frac{5}{8} \cdot \frac{1}{z+1} - \frac{1}{12} \cdot \frac{1}{z+3} + \frac{11}{24} \cdot \frac{1}{z-3} \\ &= \frac{5}{8} \cdot \frac{1}{z} \cdot \frac{1}{1 + (\frac{1}{z})} - \frac{1}{12} \cdot \frac{1}{3} \cdot \frac{1}{(\frac{z}{3}) + 1} - \frac{11}{24} \cdot \frac{1}{3} \cdot \frac{1}{1 - (\frac{z}{3})} \\ &= \frac{5}{8} \cdot \frac{1}{z} \cdot \sum_{n=0}^\infty (-1)^n\left(\frac{1}{z}\right)^{n+1} - \frac{1}{36} \cdot \sum_{n=0}^\infty \left(-\frac{z}{3} \right)^n - \frac{11}{72} \cdot \sum_{n=0}^\infty \left(\frac{z}{3} \right)^n. \end{align*}
Notice that \begin{equation*} \frac{1}{z^2+1} = \frac{i}{2} \cdot \frac{1}{z+i} - \frac{i}{2} \cdot \frac{1}{z-i}. \end{equation*} For $|z+i| < 2$, we have $\left|\frac{z+i}{2} \right| < 1$, so that \begin{align*} f(z) &= \frac{i}{2} \cdot \frac{1}{z+i} - \frac{i}{2} \cdot \frac{1}{z-i} \\ &= \frac{i}{2} \cdot \frac{1}{z+i} - \frac{i}{2} \cdot \frac{1}{(z+i) - 2i} \\ &= \frac{i}{2} \cdot \frac{1}{z+i} + \frac{i}{2} \cdot \frac{1}{2i} \cdot \frac{1}{1 - (\frac{z+i}{2i})} \\ &= \frac{i}{2} \cdot \frac{1}{z+i} + \frac{1}{4} \cdot \sum_{n=0}^\infty \left(\frac{z+i}{2i} \right)^n, \end{align*} which is valid for $0 < |z+i| < 2$.
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Am I correct? Thanks in advanced.