Laurent Series Expansion of $\frac{-3z^2+8z+1}{(z-2)(z^2+1)}$ on the annulus $A(1,2)$

I think $A(1,2)$ denotes the set $\{z:1<|z-0|<2\}$, so it excludes the poles.

using partial fraction decomposition I got;


In general The Laurent Series of $\frac{1}{z-z_0}$ about $a$

If $|z-a|<|z_0-a|$ then $\frac{1}{z-z_0}=-\frac{1}{z_0-a}\sum\limits_{n=0}^{\infty}\left(\frac{z-a}{z_0-a}\right)^n$

If $|z-a|>|z_0-a|$ then $\frac{1}{z-z_0}=\frac{1}{z-a}\sum\limits_{n=0}^{\infty}\left(\frac{z_0-a}{z-a}\right)^n$

then I have about $0$

$\frac{1}{z-2}=-\frac{1}{2}\sum\limits_{n=0}^{\infty}\left(\frac{z}{2}\right)^n\quad$ for $|z|<2$

$\frac{2}{z+i}=-\frac{2}{z}\sum\limits_{n=0}^{\infty}\left(\frac{-i}{z}\right)^n\quad$ for $|z|>1$

$\frac{2}{z-i}=-\frac{2}{z}\sum\limits_{n=0}^{\infty}\left(\frac{i}{z}\right)^n\quad$ for $|z|>1$

Can you verify my steps, and Is then their difference or sum the laurent series on the annulus, or is there another one ?


That's certainly a valid way to do it, however if I'm doing this right,$$\frac{2}{z+i} = \frac{2}{z} \cdot \frac 1 {1 - (-i/z)} = + \frac 2 z \sum_{n=0}^\infty (-i/z)^n$$which is the opposite overall sign from what you seem to have, and similarly for $2/(z - i)$. However it will be simpler to proceed if you realize that$$\frac 1 {z - i} + \frac 1 {z + i} = \frac {(z + i) + (z - i)}{(z - i)(z + i)} = \frac {2 z}{z^2 + 1}.$$This expands more simply on the annulus as:$$\frac{-4z}{z^2 + 1} = \frac{-4}{z} \cdot \frac{1}{1 - (-z^{-2})} = \frac{-4}{z} \left( 1 - z^{-2} + z^{-4} - z^{-6} + \dots\right)$$which shows more clearly that the negative even powers vanish while the rest of the negative series alternates in sign.

You can probably also get this your way; you'll get:$$i^n + (-i)^n = (i^n) \cdot (1 + (-1)^n) = (-1)^{n/2} \cdot 2 \operatorname{even}(n) $$ where $\operatorname{even}(n) = (1 + (-1)^n)/2$ is $1$ when $n$ is even and $0$ when $n$ is odd. Combined with the $2/z$ prefactor these appear to be identical, if you subtract them in keeping with the $-$ signs in your expression.

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  • $\begingroup$ Much better, thank you very much $\endgroup$ – derivative May 19 '15 at 18:25

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