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.(adsbygoogle = window.adsbygoogle || []).push({});

by now you can prob. tell im trying to get to grasp with complex analysis as a whole.

for example i have this :

find laurent series for :[itex]\[

f(z) = \frac{z}

{{z^2 - 1}}

\][/itex]

given the constraints:

(a) [itex]\[

0 < \left| {z - 1} \right| < 2

\][/itex] ............... (b) [itex]\[

\left| {z + 1} \right| > 2

\][/itex] ............... (c)[itex]\[

\left| z \right| > 1

\][/itex]

------------------------------------------------------------

My Attempt: for part (a)

first I break up the function using partial fractions:

[itex]\[

\frac{z}

{{z^2 - 1}} = \frac{z}

{{(z - 1)(z + 1)}} = \frac{A}

{{(z - 1)}} + \frac{B}

{{(z + 1)}}

\][/itex]

[itex]\[

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

\][/itex]

setting: z=1:

[itex]\[

1 = 2A,A = \frac{1}

{2}

\][/itex]

setting: z=-1:

[itex]\[

- 1 = B( - 2),B = \frac{1}

{2}

\]

[/itex]

[itex]\[

so:\frac{z}

{{z^2 - 1}} = \frac{1}

{{2(z - 1)}} + \frac{1}

{{2(z + 1)}}

\]

[/itex]

so for: 0 < |z-1| < 2:[itex]\[

\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}

\]

[/itex]

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.

-------------------------------------------------------------------------------

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Laurent Series

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**