Laurent series for 1/(z²+1) with convergence region

[krindik]

Hi,
can u pls help me on this?

Homework Statement

Find Laurent series that converges for

$$\, 0 < |z - z_0| < R }$$ and determine precise region of convergance

$$\, \frac {1}{z^2 + 1} \,\,<br />$$

Homework Equations

The Attempt at a Solution

I tried to spilt this into fractions
i.e
$$f(x) \, = \, \frac{A}{z-i} + \, \frac{B}{z+i} <br />$$

as I would have done for

$$\frac {1}{z^2 - 1} \,\,<br />$$

But in that case I would expand it with a geometrical series.
The problem rises with $$i$$ instead of $$1$$



2. Homework Statement
Can u pls explain how can I choose the method of expansion (Laurent, Taylor) given a function f(x) ?






Thanks

 

[latentcorpse]

can't you just say $w=z^2$ then you can make

$\frac{1}{w-1}=\frac{1}{w} \frac{1}{1-\frac{1}{w}}$ and expand as a geometric series...

 

[krindik]

Thanks.

In fact the question is
$$<br /> \, \frac {1}{z^2 + 1} \,\,<br />$$

not

$$<br /> \, \frac {1}{z^2 - 1} \,\,<br />$$

I'll restate the question

Find Laurent series that converges for

$$<br /> \, 0 < |z - z_0| < R }$$

and determine precise region of convergance for
$$<br /> \, \frac {1}{z^2 + 1} \,\,<br />$$

where
$$z_0 = i$$
Additionally, I would like to know how u use $$<br /> \, 0 < |z - z_0| < R }$$
in coming up with a solution

 

[Dick]

The Laurent series will be an expansion in powers of (z-i). Use your partial fractions form. The A/(z-i) is the (-1) power. Write the B/(z+i) part as B/(2i+(z-i)) and do the geometric series expansion with (z-i) as the variable.

 

[latentcorpse]

why can we not just change variable here?

 

[Dick]

You can. Isn't that basically what I did when I wrote z+i=(z-i)+2i?

 

[krindik]

Thank you,

I have few questions
1. Is it that you choose $(z - i)$ terms because the question specifies it should converge in $0 < |z-i| < R$ ?

2. So $\frac{B}{z + i}$ will be expanded like

$\frac{1}{2i} * \frac{1}{1+\frac{z-i}{2i}} = \sum_{n=0}^{\infty}{s^n}$

where
$s = \frac{z-i}{2i}$


so the answer to the solution would be
$\frac{A}{z-i} + \{expansion\, in\, 2\}$ ?

Thanks again

 

[Dick]

I chose (z-i) because you specified that you want to expand around z0=i in post 3. I couldn't tell where you wanted to put the expansion from the first post. And yes, you've got the expansion right.

 

[krindik]

Thank you very much Dick

I shall summarize here, so that anybody else following this thread can get the point.


$\frac{1}{1+z^2} = \frac{A}{z-i} + \frac{B}{z+i}$

expand
$\frac{B}{z+i} = \frac{B}{2i} * \frac{1}{1-(-1 * \frac{z-i}{2i})}$

then by expanding in geometric series
$\frac{1}{1-(-1 * \frac{z-i}{2i})} = \sum_{n=0}^\infty{[(\frac{-1}{2i})^n * (z-i)^n]}$