How can we obtain and manipulate Laurent series for different annuli?

[opticaltempest]

I uploaded a scanned page from Schaum's Outline of Complex Variables. I have some questions on how they found the Laurent series in Example 27.

http://img19.imageshack.us/img19/7172/i0001.jpg

If the image doesn't load, go to - http://img19.imageshack.us/img19/7172/i0001.jpg"

In Example 27 part (b), they used the Laurent series they obtained from part (a) and subtracted this series from the Laurent series they obtained for |z|>3 to get a Laurent series that is valid for |z|>3.

It appears that they are making substitutions into the geometric series to find the power series for 1/(2*(z+1)) (that is convergent?) for |z|>1 and the power series for 1/(2*(z+1)) (that is convergent?) for |z|>3. However, I thought the geometric series did not converge for |z|>1. How can we do these substitutions and say that we have a convergent series for |z|>1 and |z|>3? In other words, how do we know that the series for |z|>1 and |z|>3 actually converge so that we can subtract them?

 

[opticaltempest]

Well maybe since 1/(1-z) converges for |z|>1, with z=1/z, we get |1/z|<1 which is equivalent to |z|>1.

Is multiplying the fraction by 1/z a standard procedure for finding the series that converges for |z|>1 ?

 

[n!kofeyn]

The geometric series is \frac{1}{1-z} = \sum_{n=0}^\infty z^n and converges for |z|&lt;1. It diverges for |z|&gt;1.

If |z|&gt;1, then |1/z|&lt;1. The way they get the series expansion in part (a) is:
\frac{1}{2(z+1)} = \frac{1}{2z} \cdot \frac{1}{1-\left(-\frac{1}{z} \right)} = \frac{1}{2z} \sum_{n=0}^\infty \left( -\frac{1}{z} \right)^n = \frac{1}{2z} \sum_{n=0}^\infty (-1)^n \frac{1}{z^n} = \sum_{n=0}^\infty (-1)^n \frac{1}{2z^{n+1}} = \frac{1}{2z} - \frac{1}{2z^2} + \frac{1}{2z^3} - \cdots


Now in part (b), if |z|&gt;3, then |1/z|&lt;3, and thus |3/z|&lt;1. Then again using the geometric series:
\frac{1}{2(z+3)} = \frac{1}{2z}\cdot \frac{1}{1-\left( -\frac{3}{z} \right)} = \frac{1}{2z} \sum_{n=0}^\infty \left( -\frac{3}{z} \right)^n = \frac{1}{2z} \sum_{n=0}^\infty (-1)^n \frac{3^n}{z^n} = \sum_{n=0}^\infty (-1)^n \frac{3^n}{2z^{n+1}} = \frac{1}{2z} - \frac{3}{2z^2} + \frac{9}{2z^3} - \cdots


For part (b), we need the Laurent expansion for the annulus 3&lt;|z|&lt;\infty. (Laurent expansions are always defined for an annulus.) Both of the series we have above converge for |z|>3. (The first converges for |z|&gt;1, which means it converges for |z|&gt;3.) Then we can just add the series terms together to get the final Laurent expansion in the annulus 3&lt;|z|&lt;\infty.

Hope this helps!