# Laurent Expansion about 0, valid in $1 < \vert z \vert < 2$, complicated function with complex coefficients

So, I have a homework question about Laurent series expansion - I feel comfortable with these question but this one is quite a bit more complicated and I wanted to ask whether what I've done makes sense for a function like this? So, I need to find a Laurent expansion about $$0$$, valid in the annulus $$1 < \vert z \vert < 2$$ for

$$f(z) = \frac{z+1}{(z+2)(z-i)}$$

1. Partial fraction decomposition:

$$f(z) = \frac{z+1}{(z+2)(z-i)} = \frac{\frac{2}{5}-\frac{1}{5}i}{z+2} + \frac{\frac{3}{5}+\frac{1}{5}i}{z-i}$$

1. Find series valid for $$\vert z \vert > 1$$

\begin{align} \frac{\frac{3}{5}+\frac{1}{5}i}{z-i} &= i\left(\frac{3}{5}+\frac{1}{5}i\right)\left(\frac{1}{1 - zi}\right)\\ \\ &= i\left(\frac{3}{5}+\frac{1}{5}i\right)\sum_{n = -\infty}^{-1}(-zi)^{n}, \; \vert zi\vert > 1\\ \\ &= \sum_{n = -\infty}^{-1}\left(\frac{3}{5}+\frac{1}{5}i\right)(-1)^{n}z^{n}i^{n+1}, \; \vert z\vert > 1 \end{align}

1. Find series valid for $$\vert z \vert < 2$$

\begin{align} \frac{\frac{2}{5}-\frac{1}{5}i}{z+2} &= 2\left(\frac{2}{5}-\frac{1}{5}i\right)\left(\frac{1}{1 - (-z/2)}\right)\\ \\ &= 2\left(\frac{2}{5}-\frac{1}{5}i\right)\sum_{n = 0}^{\infty}\left(-\frac{z}{2}\right)^{n}, \; \left| -\frac{z}{2}\right| < 1\\ \\ &= \sum_{n = 1}^{\infty}\left(\frac{2}{5}-\frac{1}{5}i\right)(-1)^{n -1}\frac{z^{n-1}}{2^{n}}, \; \left| z\right| < 2 \end{align}

1. Thus, \begin{align} f(z) = \frac{z+1}{(z+2)(z-i)} = \sum_{n = -\infty}^{-1}\left(\frac{3}{5}+\frac{1}{5}i\right)(-1)^{n}z^{n}i^{n+1} + \sum_{n = 1}^{\infty}\left(\frac{2}{5}-\frac{1}{5}i\right)(-1)^{n-1}\frac{z^{n-1}}{2^{n}}\\ \end{align}

Valid in the annulus $$1 < \vert z \vert < 2$$

Does this work?

• Aren't there some signs problems in 2? $\frac1{z-i}$ is not $\frac i{1-zi}$ but $\frac i{1+zi},$ and $\lim_{|z|\to\infty}\frac z{z-i}=1\ne-1=\lim_{|z|\to\infty}z\sum_{n = -\infty}^{-1}(-1)^{n}z^{n}i^{n+1}.$ Commented Nov 26, 2022 at 13:55
• @AnneBauval Ah, yes, thank you. So it would be $\sum_{-\infty}^{-1}\left(\frac{3}{5} + \frac{1}{5}i\right)z^{n}i^{n+1}$ instead, yes? Commented Nov 26, 2022 at 14:50
• I found $\sum_{-\infty}^{-1}\left(\frac{3}{5} + \frac{1}{5}i\right)z^{n}(-i)^{n+1}.$ Commented Nov 26, 2022 at 14:57
• Like this? \begin{align} \frac{\frac{3}{5}+\frac{1}{5}i}{z-i} &= i\left(\frac{3}{5}+\frac{1}{5}i\right)\left(\frac{1}{1-((-z)i)}\right)\\ &= i\left(\frac{3}{5}+\frac{1}{5}i\right)\sum_{n = -\infty}^{-1}(-(-z)i)^{n}\\ &= i\left(\frac{3}{5}+\frac{1}{5}i\right)\sum_{n = -\infty}^{-1}z^n(-i)^n \end{align} $i\left(\frac{3}{5}+\frac{1}{5}i\right)\sum_{n = -\infty}^{-1}z^n(-i)^n = \left(\frac{3}{5}+\frac{1}{5}i\right)\sum_{n = -\infty}^{-1}z^n(-i)^{n+1}$ ? Commented Nov 26, 2022 at 15:19
• Again, $\lim_{|z|\to\infty}\frac z{z-i}=1\ne-1=\lim_{|z|\to\infty}zi\sum_{n = -\infty}^{-1}z^n(-i)^n.$ Commented Nov 26, 2022 at 15:59

Let us forget about $$\frac{3+i}5.$$ For $$|z|>1,$$ \begin{align}\frac1{z-i}&=\frac1z\frac1{1-\frac iz}\\&=\frac1z\sum_{n\ge0}\left(\frac iz\right)^n \\&=\frac1z\sum_{k\le0}(-iz)^k \\&=\frac1z\sum_{n\le-1}(-iz)^{n+1}\\&=\sum_{n\le-1}z^n(-i)^{n+1}.\end{align}