Expand $f(z) = \frac{1}{z(z-1)(z-2)}$ in the region $0 < |z| < 1$:

Using partial fraction decomposition, $f(z) = \frac{1}{2} \cdot \frac{1}{z}-\frac{1}{z-1} + \frac{1}{2}\cdot\frac{1}{z-2}$.

So, using some algebra I have

$f(z) = \frac{1}{2} \cdot \frac{1}{z}+\frac{1}{1-z} - \frac{1}{4}\cdot\frac{1}{1-\frac{z}{2}} = {\frac{1}{2} \cdot \frac{1}{z} + \sum_{n=0}^\infty z^n -\frac{1}{4}\sum_{n=0}^\infty (\frac{z}{2})^n}$.

Also, I need to expand in the region $1 < |z| < 2$. How does this change the problem? The only thing I think I have to do different is rewrite the $-\frac{1}{z-1}$ term as $-\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}$, but I am not entirely sure.

Any help appreciated.


Actually, it's better to use the partial fraction decomposition$$\frac1{(z-1)(z-2)}=-\frac1{2-z}+\frac1{1-z}$$and to divide everything by $z$ at the end.

So, if $0<\lvert z\rvert<1$, then$$-\frac1{2-z}+\frac1{1-z}=-\frac12\cdot\frac1{1-\frac z2}+\frac1{1-z}=-\frac12\sum_{n=0}^\infty\frac{z^n}{2^n}+\sum_{n=0}^\infty z^n.\tag1$$Therefore, the Laurent series of $f$ is $(1)$ divided by $z$.

If $1<\lvert z\rvert<2$, then$$-\frac1{2-z}+\frac1{1-z}=-\frac12\cdot\frac1{1-\frac z2}+\frac1{1-z}=-\frac12\sum_{n=0}^\infty\frac{z^n}{2^n}-\sum_{n=-\infty}^{-1}z^n.\tag2$$Again, the Laurent series of $f$ is $(2)$ divided by $z$.

  • $\begingroup$ I see. I'll probably use this, but was my original logic incorrect? For the second region could I write the $\frac{1}{z-1}$ term as $\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}$, which is valid in the region since $|z| > 1$ means that $|\frac{1}{z}| < 1$? $\endgroup$ – mXdX Apr 11 at 14:55
  • 1
    $\begingroup$ What you did was correct. I just think that it is simplesr to divide by $z$ in the end. And, yes, It is fine, in the second region to use the equality $\frac1{z-1}=\frac1z\cdot\frac1{1-\frac1z}$ and the fact that $\lvert\frac1z\rvert<1$. $\endgroup$ – José Carlos Santos Apr 11 at 14:57

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