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 $$|z|<1$$. It diverges for $$|z|>1$$.

If $$|z|>1$$, then $$|1/z|<1$$. 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 $$|z|>3$$, then $$|1/z|<3$$, and thus $$|3/z|<1$$. 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 $$3<|z|<\infty$$. (Laurent expansions are always defined for an annulus.) Both of the series we have above converge for |z|>3. (The first converges for $$|z|>1$$, which means it converges for $$|z|>3$$.) Then we can just add the series terms together to get the final Laurent expansion in the annulus $$3<|z|<\infty$$.

Hope this helps!