Example 7.7.  Find three different Laurent series representations for the function    involving powers of z.

Explore Solution 7.7.

We could let Mathematica sum up the various series, and note when they are valid.

Combining the above representations we get:

If more details are desired then we can proceed here they are.

Enter the function   and find the first few terms of the various Laurent series.

Mathematica easily finds the Maclaurin series involving positive powers of  z.

The above series converges for  ,  this requires looking at the sum of the two "geometric series" that form the parts of  .  Mathematica can also find "the other series" that involves negative powers of  z, this involves the mental thinking that you substitute    in the original series to get a new function of  Z, then expand it about  Z = 0  and then substitute  .  

The above series converges for  ,  this requires looking at the sum of the two "geometric series that form the parts of .
This leaves an unresolved question: "What happens in the annulus   ?  This requires a little work.  First split up the functions up into their partial fraction form, and then do expansions about    for each part.

The singularity of  ,  and the singularity of  .  Next we form Laurent expansions for the two parts  ,  there are two expansions for each function.

Laurent Series (i).  The first Laurent series    is valid for  .  

We can use Mathematica to investigate how well the Laurent series is "converging" for real numbers.

We can use Mathematica to investigate how well the complex Laurent series is "converging" in the complex plane.
For the first series we need to be close to  .  For convenience we use the disk  .  

Aside.  If z is not chosen close enough to then the series will not converge rapidly near the circle.  More terms are needed to get convergence near the circle  .  We can see what is happening in the disk  .  

Laurent Series (ii).  The second Laurent series    is valid for  .  

We can use Mathematica to investigate how well the Laurent series is "converging" for real numbers.

We can use Mathematica to investigate how well the complex Laurent series is "converging" in the complex plane.
For the second series we need to be in the middle of the annulus  .   For convenience we choose  .  

Hence is an approximation to    valid for  .   
Notice that the first series    is not a good approximation in  ,  for illustration, it is not accurate at  .  

Laurent Series (iii).  The third Laurent series    is valid for  .  

We can use Mathematica to investigate how well the Laurent series is "converging" for real numbers.

We can use Mathematica to investigate how well the complex Laurent series is "converging" in the complex plane.
For the third series z must satisfy  .  For convenience we choose  .  

Aside. If z is not large enough to then the series will not converge rapidly.  More terms are needed to get convergence near  .    We can see what is happening in the annulus  .  

So    is an approximation to    valid for  .  
Notice that are not good approximations for  ,  for illustration they are not accurate at  z = 4.0

Divergent Series Combination (iv).  Aside.  The fourth possibility    does not converge for any values of z and should not be considered, however you could write down a few terms and think about why this series diverges.  

In the above series, when    the portion of the series with positive exponents will converge but the portion with negative exponents will diverge. Similarly, when    the portion of the series with negative exponents will converge, but the portion of the series with positive exponents will diverge.   Hence, the series diverges for all  z.  It does not approximate  f[z]  at any of the values x = 0.5, 1.5, 4.0

We can summarize the above findings.

(c) 2006 John H. Mathews, Russell W. Howell