Recent content by yakeyglee

Take note of the following. f(z) = \frac{1}{z^2 - 4} = \frac{1}{z-2} \cdot \frac{1}{z+2} = \frac{1}{z - 2} \cdot \frac{\frac{1}{4}}{1 - \left( - \frac{z - 2}{4} \right)} = \frac{1}{z-2} \cdot \frac{1}{4} \sum_{n=0}^\infty \left( - \frac{z - 2}{4} \right)^n = \cdots