I would really appreciate help finding the Laurent series for the function $f(z)=\frac{z}{z-i}+\frac{z}{z+4}$ at :
a) 0<|z|<1
b)1<|z|<4
c) 0<|z-i|<√(17)
Here are my best attempts but I'm honestly not confident any is right. I'd really appreciate any help someone could offer
a) $\frac{z}{z-i}$= $\frac{-1}{i}\frac{1}{1-\frac{z}{i}}$=$i\frac{1}{1+zi}$=$\sum_{j=0}^∞ (-1)^j z^j i^{j+1}$
$\frac{z}{z+4}$= $\frac{1}{4}\frac{1}{1+\frac{z}{4}}$=$\sum_{j=0}^∞ (-1)^j\frac{z^j}{4^{j+1}}$
--> $z[\sum_{j=0}^∞ (-1)^j z^j i^{j+1} + \sum_{j=0}^∞ (-1)^j\frac{z^j}{4^{j+1}}$]= $\sum_{j=1}^∞ (-1)^j{z^j}(i^j + \frac{1}{4^{j+1}})$
Then for b) I'm not sure how to change anything from a)..
c)$\frac{1}{z+4}$=$\frac{1}{(z-i)+4}$=$\frac{1}{z-i}\frac{1}{1+\frac{4}{z-i}}$=$\frac{1}{z-1}\sum_{j=0}^∞ \frac{4}{(z-i)^{j}}$
--> $z[\frac{1}{z-1}][\frac{1}{z-1}\sum_{j=0}^∞ \frac{4}{(z-i)^{j}}$]=$z\sum_{j=0}^∞ \frac{4}{(z-i)^{j+2}}$
Part (b): The function $\frac{z}{z-i}$ is holomorphic in $|z| > 1$ and in $|z| < 1$, and the function $\frac{z}{z+4}$ is holomorphic in $|z| > 4$ and in $|z| < 4$. So you want the expansion of the first function in $|z| > 1$, and the expansion of the second in $|z| < 4$. $$ \frac{z}{z+4} = \frac{z}{4}\frac{1}{1+\frac{z}{4}} = \frac{z}{4}\sum_{n=0}^{\infty}(-1)^{n}\frac{z^{n}}{4^{n}},\;\;\; |z| < 4. $$ The remaining function must be expanded in $|z| > 1$: $$ \frac{z}{z-i}=\frac{1}{1-\frac{i}{z}}=\sum_{n=0}^{\infty}\frac{i^{n}}{z^{n}},\;\;\; |z| > 1 $$ Part (c): The function $\frac{z}{z-i}$ is holomorphic in $0 < |z-i|$. The function $\frac{z}{z+4}=\frac{z}{(z-i)+(4+i)}$ is holomophic in $|z-i| < |4+i|=\sqrt{17}$. $$ \frac{z}{z-i}=\frac{z-i+i}{z-i}=1+\frac{i}{z-i} \\ \frac{z}{z+4} = \frac{z}{(z-i)+(4+i)}=\frac{(z-i)+i}{4+i}\frac{1}{1+\frac{z-i}{4+i}} $$