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I need to find Laurent expansion for function $$f(z) = \frac{iz^2 + 4iz + 4 +12 i}{(z^2+4)(z+2-i)}$$ for $2 < |z-i| < 3$. I start with division: $$\frac{iz^2 + 4iz + 4 +12 i}{(z^2+4)(z+2-i)} = \frac{1}{z-2i} - \frac{1}{z+2i} + \frac{i}{z+2-i}$$ Then I find Taylor expansions of these fractions: $$\frac{1}{z-2i} = \sum_{n=0}^\infty \frac{1}{n!} (z-2i)^{(n)}(i) (z-i)^n$$ $$\frac{1}{z+2i} = \sum_{n=0}^\infty \frac{1}{n!} (z+2i)^{(n)}(i) (z-i)^n$$ $$\frac{1}{z+2i} = \sum_{n=0}^\infty \frac{1}{n!} (z+2-i)^{(n)}(i) (z-i)^n$$ and I am supposed to combine those into Laurent expansion, but I don't know how procced from sum of Taylor expansions to Laurent expansion. How do I do that?

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The first series you found is not valid in the desired region $2<|z-i|<3$, but rather $|z-i|<1$: $$\begin{align*} \frac{1}{z-2i}&=\frac{i}{1-\frac{z-i}{i}}\\[1ex] &=i\sum_{n=0}^\infty \left(\frac{z-i}{i}\right)^n\\[1ex] &=\sum_{n=0}^\infty i^{1-n}(z-i)^n \end{align*}$$

You want a series valid outside of this region ($|z-i|>1$ contains $2<|z-i|<3$), but the Taylor expansion won't allow for it in this form.

By some algebraic manipulation, we can instead write $$\begin{align*} \frac{1}{z-2i}&=\frac{1}{(z-i)-i}\\[1ex] &=\frac{1}{z-i}\times\frac{1}{1-\frac{i}{z-i}}\\[1ex] &=\frac{1}{z-i}\sum_{n=0}^\infty \left(\frac{i}{z-i}\right)^n\\[1ex] &=\sum_{n=0}^\infty i^n (z-i)^{-(n+1)} \end{align*}$$ which is valid for $\left|\dfrac{1}{z-i}\right|<1$, or equivalently, $|z-i|>1$. This is the Laurent expansion for the first partial fraction term.

See if you can adapt this procedure to the other two terms.

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  • $\begingroup$ Can you explain difference between first and second expansions, and why first one is not vaild? I fail to understand the difference. $\endgroup$ Jun 3, 2016 at 16:47
  • $\begingroup$ A geometric series $\sum\limits_{n\ge0}r^n$ converges only for $|r|<1$, whereas $\sum\limits_{n\ge0}\left(\frac{1}{r}\right)^n$ converges for $|r|>1$. The idea is that the second form - the actual Laurent expansion - allows you to write a series expansion outside the usual region of convergence for geometric series (the unit disk centered at $r=0$ in this case). You can read up more on the distinction in the answer here. $\endgroup$
    – user170231
    Jun 3, 2016 at 17:06
  • $\begingroup$ Thanks, I think I understand now. $\endgroup$ Jun 3, 2016 at 17:11

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