2. Expand $\frac{7}{(z- 2)(z+5)^2}$ as $\frac{\frac{1}{7}}{z- 2}+ \frac{\frac{5}{14}}{z+ 5}- \frac{1}{(z+5)^2}$ using partial fractions.
Now expand $\frac{\frac{1}{7}}{z-2}$ as a power series in z+ 5. You can do that, for example, by writing it as $\frac{\frac{1}{7}}{z+ 5- 7}= -\frac{\frac{1}{7}}{7- (z+5)}$ $= -\frac{\frac{1}{49}}{1- \frac{z+ 5}{7}}$ and recognize that as the sum of a geometric series: $\sum_{n=0}^\infty-\frac{1}{49}\left(\frac{z+5}{7}\right)^n$. The Laurent series is $-(z+5)^{-2}+ \frac{5}{14}(z+5)^{-1}+ \sum_{n=0}^\infty-\frac{1}{49}\left(\frac{z+5}{7}\right)^n$