Laurent series of f(z) Can you check my calculations?
Find Laurent series for: $f(z)=\frac{6z+8}{(2z+3)(4z+5)}=\frac{1}{4z+5}+\frac{1}{2z+3}$
when a)$\frac{5}{4}<|z|<\frac{3}{2}$
b)$|z|>\frac{3}{2}$
a)$\sum_{n=0}^\infty \frac{(-2)^nz^n}{3^{n+1}} + \sum_{n=1}^\infty \frac{(-4)^nz^{-n}}{5^{n+1}}$
b)$\sum_{n=0}^\infty \frac{(-2)^nz^{-n}}{3^{n+1}} + \sum_{n=1}^\infty \frac{(-4)^nz^{-n}}{5^{n+1}}$
Is it important which series I will take since n=1? Thanks in advance

Note: Regarding your question, the series we take is important. Here I provide a complete Laurent expansion around $z=0$. From this you should be able to check our results (which slightly differ). 

The function
\begin{align*}
 f(z)&=\frac{1}{4}\frac{1}{z+\frac{5}{4}}+\frac{1}{2}\frac{1}{z+\frac{3}{2}}\\
\end{align*}
   has two simple poles at $-\frac{5}{4}$ and $-\frac{3}{2}$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $-\frac{5}{4}$ and $-\frac{3}{2}$ and see they determine three regions.
\begin{align*}
 |z|<\frac{5}{4},\qquad\quad
 \frac{5}{4}<|z|<\frac{3}{2},\qquad\quad
 \frac{3}{2}<|z|
 \end{align*} 
  
  
*
  
*The first region $ |z|<\frac{5}{4}$ is a disc with center $0$, radius $\frac{5}{4}$ and the pole $-\frac{5}{4}$ at the boundary of the disc. In the interior of this disc all two fractions with poles $-\frac{5}{4}$ and $-\frac{3}{2}$  admit a representation as power series at $z=0$.
  
*The second region $\frac{5}{4}<|z|<\frac{3}{2}$ is the annulus with center $0$, inner radius $\frac{5}{4}$ and outer radius $\frac{3}{2}$. Here we have a representation of the fraction with pole $\frac{5}{4}$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $\frac{3}{2}$ admits a representation as power series.
  
*The third region $|z|>\frac{3}{2}$ containing all points outside the disc with center $0$ and radius $\frac{3}{2}$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.

A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}

We can now obtain the Laurent expansion of $f(x)$ at $z=0$ for all three regions
  
  
*
  
*Region 1: $|z|<\frac{5}{4}$
  
  
  \begin{align*}
f(z)&=\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{4}{5}\right)^{n+1}(-z)^n
+\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^{n+1}(-z)^n\\
&=\sum_{n=0}^{\infty}\left(\frac{4^n}{5^{n+1}}+\frac{2^{n}}{3^{n+1}}\right)(-z)^n
\end{align*}
  
  
*
  
*Region 2: $\frac{5}{4}<|z|<\frac{3}{2}$
  
  
  \begin{align*}
f(z)&=-\frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{5}{4}\right)^{n-1}\frac{1}{(-z)^n}
+\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^{n+1}(-z)^n\\
&=-\sum_{n=1}^{\infty}\frac{5^{n-1}}{4^n}\frac{1}{(-z)^n}+\sum_{n=0}^{\infty}\frac{2^n}{3^{n+1}}(-z)^n
\end{align*}
  
  
*
  
*Region 3: $\frac{3}{2}<|z|$
  
  
  \begin{align*}
f(z)&=-\frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{5}{4}\right)^{n-1}\frac{1}{(-z)^n}
-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{3}{2}\right)^{n-1}\frac{1}{(-z)^n}\\
&=-\sum_{n=1}^{\infty}\left(\frac{5^{n-1}}{4^n}+\frac{3^{n-1}}{2^n}\right)\frac{1}{(-z)^n}
\end{align*}

