I have to find the Laurent-Series with $0<|z|<5/2$ and $z_0=0$ for: $$\frac{8z+6+8i}{2z^2-3z-4iz}$$
I already did this: $$ \frac{8z+6+8i}{2z^2-3z-4iz} = \frac{A}{2\cdot(z-0)}+ \frac{B}{2\cdot(z-1.5-2i)}\\ A=-2 \text{ and } B=6 \\ \frac{8z+6+8i}{2z^2-3z-4iz} = -1\cdot\frac{1}{z} + 3\cdot\frac{1}{z-1.5-2i} $$ I hope this is correct so far. My book says the formula for a Laurent Series is : $$\sum_{k=1}^{\infty} a_k\cdot\frac{1}{[z-z_0]^k} + \sum_{k=0}^{\infty}b_k\cdot[z-z_0]^k=\sum_{k=-\infty}^{\infty} c_{k}\cdot[z-z_{0}]^{k}$$
But I have no Idea what i have to do now.
I hope someone can help me.
Thank you!
You are on the right track. After the decomposition (please check your computations), $$\frac{8z+6+8i}{2z^2-3z-4iz}=-\frac{2}{z}+\frac{12}{2z-(3+4i)}= -\frac{2}{z}-\frac{\frac{12}{3+4i}}{1-\frac{2z}{(3+4i)}}$$ you should expand $(1-\frac{2z}{(3+4i)})^{-1}$ at $z=0$ (note that it is holomorphic in the disc $|z|< |(3+4i)/2|=\sqrt{2^2+4^2}/2=5/2$) as $$\sum_{k=0}^{\infty}\left(\frac{2z}{(3+4i)}\right)^k.$$ Can you take it from here?