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• (가)$\displaystyle \frac{z+1} {\left (z-1 \right) \left (z+5 \right)} = \sum _ {n=0} ^ {\infty} \frac{1} {3} \left [\frac{2 \left (-1 \right) ^n} {5^ {n+1}} -1 \right] z^n$, $0 < \left | z \right | <1$
• (나)$\displaystyle \frac{3z-1} {\left (z-2 \right) \left (z+3 \right)} = \sum _ {n=1} ^ {\infty} [2^ {n-1} +2 \left (-3 \right) ^ {n-1}] z^ {-n}$, $\left | z \right | >3$
• (다)$\displaystyle \frac{2 \left (z+2 \right)} {z^2 -1} = - \sum _ {n=1} ^ {\infty} \frac{\left (z+2 \right) ^n} {3^n} - \sum _ {n=0} ^ {\infty} \left (z+2 \right) ^ {-n}$, $1 < \left | z+2 \right | <3$
• (라)$\displaystyle \begin{aligned} \frac{10} {\left (z+2 \right) \left (z ^ {2} +1 \right)} \\ &= \sum _ {n=0} ^ {\infty} \left (- \frac{1} {2} \right) ^ {n} z ^ {n} + \sum _ {n=0} ^ {\infty} 2 \left (-1 \right) ^ {n+1} z ^ {- (2n+1)} + \sum _ {n=1} ^ {\infty} 4 \left (-1 \right) ^ {n+1} z ^ {-2n} \end{aligned}$
• , $1< \left | z \right | <2$