Originally Posted by

**mathfied** find laurent series for :$\displaystyle

f(z) = \frac{z}

{{z^2 - 1}}

$

given the constraints:

(a) $\displaystyle

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

$ ............... (b) $\displaystyle

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

$ ............... (c)$\displaystyle

\left| z \right| > 1$

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

first I break up the function using partial fractions:

...

$\displaystyle

\frac{z}

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

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

{{2(z + 1)}}

$

**so for: 0 < |z-1| < 2:**

$\displaystyle

\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?