# Exploring Laurent Expansions with Constraints

• mathfied

#### mathfied

i get the idea of the laurent expansion but i get confused with the constraints and how they change the way you work with the expansion.
by now you can prob. tell I am trying to get to grasp with complex analysis as a whole.

for example i have this :

find laurent series for :$$f(z) = \frac{z} {{z^2 - 1}}$$

given the constraints:
(a) $$0 < \left| {z - 1} \right| < 2$$ ... (b) $$\left| {z + 1} \right| > 2$$ ... (c)$$\left| z \right| > 1$$

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My Attempt: for part (a)

first I break up the function using partial fractions:
$$\frac{z} {{z^2 - 1}} = \frac{z} {{(z - 1)(z + 1)}} = \frac{A} {{(z - 1)}} + \frac{B} {{(z + 1)}}$$

$$z = A(z + 1) + B(z - 1)$$

setting: z=1:
$$1 = 2A,A = \frac{1} {2}$$

setting: z=-1:
$$- 1 = B( - 2),B = \frac{1} {2}$$

$$so:\frac{z} {{z^2 - 1}} = \frac{1} {{2(z - 1)}} + \frac{1} {{2(z + 1)}}$$

so for: 0 < |z-1| < 2:
$$\begin{gathered} \frac{1} {{2(z - 1)}} + \frac{1} {{2(z + 1)}} \hfill \\ = \frac{1} {{2(z - 1)}} + \frac{1} {2}\left[ {\frac{1} {{2 - ( - (z - 1)}}} \right] \hfill \\ = \frac{1} {{2(z - 1)}} + \frac{1} {2}\frac{1} {2}\left[ {\frac{1} {{1 - \left[ { - (\frac{{z - 1}} {2})} \right]}}} \right] \hfill \\ = \frac{1} {{2(z - 1)}} + \frac{1} {4}\left[ {\frac{1} {{1 - \left[ { - (\frac{{z - 1}} {2})} \right]}}} \right] \hfill \\ = \frac{1} {{2(z - 1)}} + \frac{1} {4}\sum\limits_{n = 0}^\infty {\left[ {( - 1)^n \frac{{(z - 1)^n }} {{2^n }}} \right]} \hfill \\ \frac{1} {{2(z - 1)}} + \sum\limits_{n = 0}^\infty {\left[ {( - 1)^n \frac{{(z - 1)^n }} {{2^{n + 2} }}} \right]} \hfill \\ \end{gathered}$$

is this correct? I'm predicting that i may have missed out two parts:
1) the first part of the final equation : 1/2(z-1) : can this be simplified and integrated into the sum formula?

2) the constraint for part (a) was 0 < |z-1| < 2. I didn't know how to interpret |z-1| being between 0 and 2.

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for the other two parts - the constraints are (part (b) |z+1|>2 , part (c) |z|>1) - please could you advise me on how the constraints are meant to be used and how the final answer changes? is there a trick to this?

## Answers and Replies

For part (b), the Laurent series for your function is:$f(z) = \frac{z}{{z^2 - 1}} = \frac{A}{(z+1)} + \frac{B}{(z-1)} = \frac{1}{2(z+1)} + \frac{1}{2(z-1)}$where A and B are constants. This can be written as:$f(z) = \frac{1}{2(z+1)} + \frac{1}{2}\left[{\frac{1}{2-(-(z-1))}}\right] = \frac{1}{2(z+1)} + \frac{1}{2}\frac{1}{2}\left[{\frac{1}{1-(-(\frac{z-1}{2}))}}\right]$and then simplified to give:$f(z) = \frac{1}{2(z+1)} + \sum_{n=0}^{\infty}\left[{(-1)^n\frac{(z-1)^n}{2^{n+2}}}\right]$For part (c), you can simply substitute |z| > 1 into the above equation, giving:$f(z) = \frac{1}{2(z+1)} + \sum_{n=0}^{\infty}\left[{(-1)^n\frac{|z|^n}{2^{n+2}}}\right]$