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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!

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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?

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  • $\begingroup$ Thanks for your fast answer. I am completly confused now. I thought I would take the two parts (the A and he B part) and bring them into the form of of the formular? But why ignore $-\frac{2}{z}$ and only go with the B part? Why change the second part? I thought 1/(z-something) would be the right approach to get it into the formular. Sorry I really don't understand whats happening here. $\endgroup$ – kitty Jan 27 '19 at 18:54
  • $\begingroup$ No, you have to consider also $2/z$ which is ready for the Laurent expansion. Note that here $z_0=0$ so the Laurent expansion has the form $\sum_{k=1}^{\infty}\frac{a_k}{z^k} + \sum_{k=0}^{\infty}b_k\cdot z^k$. $\endgroup$ – Robert Z Jan 27 '19 at 19:06

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