Laurent Expansion

**mathfied** · July 9th 2008, 09:54 AM

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 im 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:
$ \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] $

$ = \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]} $

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?

**Opalg** · July 9th 2008, 12:33 PM

Originally Posted by mathfied

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$

------------------------------------------------------------
My Attempt: for part (a)

first I break up the function using partial fractions:

...

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

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

is this correct? That looks completely correct to me.

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? No, it doesn't need simplifying. It represents the one negative power of z-1 in that expansion.

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. The reason that this is the correct expansion in this region is that the condition for the infinite series to converge is |(z-1)/2| < 1.

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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?

For part (b), you want the series to converge when |z+1|>2, or in other words |2/(z+1)| < 1. So this time, you need to get an expansion in terms of negative powers of z+1.

Take the same partial fractions expression as before, but write it as

$\frac1{2(z + 1)} + \frac1{2(z - 1)} = \frac1{2(z + 1)} + \frac12\left[\frac1{(z+1)-2}\right] = \frac1{2(z + 1)} + \frac1{2(z+1)}\left[\frac1{1-\frac2{z+1}}\right].$

Then expand the expression in square brackets using the binomial series.

For (c), go back to the original expression $f(z) = \frac{z}{z^2 - 1}$ . To expand this in powers of z, you would need a series in positive powers of z if you wanted it to converge when |z| < 1. But the question asks for an expansion valid when |z| > 1, so you must use negative powers of z. Write the function as $f(z) = \frac1z\left[\frac1{1-\frac1{z^2}}\right],$ and again use the binomial expansion of the expression in the square brackets.

**mathfied** · July 11th 2008, 07:58 AM

for the second one (part b), i got:

$ \frac{1} {{2(z + 1)}} + \frac{1} {{2(z + 1)}}\left[ {\frac{1} {{1 - \frac{2} {{z + 1}}}}} \right] = \frac{1} {{2(z + 1)}} + \frac{1} {{2(z + 1)}}\sum\limits_{n = 0}^\infty {\left[ {\frac{{2^n }} {{(z + 1)^n }}} \right]} =$
$ \frac{1} {{2(z + 1)}} + \sum\limits_{n = 0}^\infty {\left[ {\frac{{2^{n - 1} }} {{(z + 1)^{n + 1} }}} \right]} $

would this be correct?

ill now proceed to post the final part..

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