Problem of class 38

For

$$f(z)=\frac{1}{(z-i)(z+i)}$$

consider $f(z)=\frac{A}{z-i}+\frac{B}{z+i}$ and solve $A,B$
And it’s $f(z)=\frac{-i/2}{z-i}+\frac{i/2}{z+i}$.
Then do Power theory:

$$f(z)=\frac{1}{2}\left(\frac{1}{1-(\frac{-z}{i})}+\frac{1}{1-\frac{z}{i}}\right)=\frac{1}{2}\left[\sum _{k=0}^{\infty }(\frac{-z}{i}{)}^k+\sum _{k=0}^{\infty }(\frac{z}{i}{)}^k\right]=\sum _{k=0}^{\infty }\frac{z^{2k}}{i^{2k}}$$

$\square$