Laurent Series (Complex Analysis)

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HansBu
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Homework Statement
Laurent Series (Complex Analysis)
Relevant Equations
f(z) = An(z-z_o)^n
My homework is on mathematical physics and I want to know the concept behind Laurent series. I want to know clearly know the process behind attaining the series representation for the expansion in sigma notation using the formula that can be found on the attached files. There are three questions and I hope you will be able to help me in this subject matter. Thank you and God bless! This was retrieved from Charlie Harper's Analytic Method in Physics . Below is the formula needed, the problems, and the attempt to the solution.

On (a), I checked my answer with wolfram alpha however, my series representation (sigma notation) is wrong. Also in number 2, i got it wrong by using simple concept of geometric series. I believe item (c) is correct. Are my answers correct already?
 
Attachments
  • Textbook problem set asking for Laurent expansions: (a) 1/(Z(1-Z)^2) about Z0=0 and Z0=1; (b) e^Z/(Z-2)^3 about Z0=3; (c) 1/((Z-1)(Z-2)) in the region |Z|<1, with a hint to use partial fractions.
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  • Two displayed equations from a complex analysis text: the Laurent series f(Z) = sum from n = -infinity to infinity of A_n (Z - Z_0)^n, and A_n = (1/(2 pi i)) times the contour integral over C of f(xi) d xi divided by (xi - Z_0)^(n+1), for n = 0, plus or minus 1, plus or minus 2, and so on.
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  • Handwritten notes on complex Laurent series: the general Laurent expansion formula with analytic and principal parts, followed by worked expansions of 1/((z-1)(z-2)), e^z/(z-3)^3 and 1/(z(1-z)^2) as sigma-notation series.
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Please post your working in typed form. It is easier to read and easier to reference when making comments.
It looks like your answer to a), z0=0 is ##\Sigma_0^\infty (-n+1)z^{n-1}##. Am I reading that right? It does not seem to follow from the preceding line.
 
haruspex said:
Please post your working in typed form. It is easier to read and easier to reference when making comments.
It looks like your answer to a), z0=0 is ##\Sigma_0^\infty (-n+1)z^{n-1}##. Am I reading that right? It does not seem to follow from the preceding line.
It doesn't have a negative sign on it. The one about z_o = 1, I think it's wrong because I validated it using WolframAlpha.
 
HansBu said:
It doesn't have a negative sign on it.
Then it's an example of why the forum rules say that images are for textbook extracts and diagrams.
 
HansBu said:
The one about z_o = 1, I think it's wrong because I validated it using WolframAlpha.
When you introduced the (-1)n factor at the end, did you mean to switch the 1-z to z-1?
 
haruspex said:
When you introduced the (-1)n factor at the end, did you mean to switch the 1-z to z-1?
Yes. That is what I meant. Is the math wrong?
 
HansBu said:
Yes. That is what I meant. Is the math wrong?
What I mean is, you added the alternating sign factor but did not switch 1-z to z-1, so yes, the last line is wrong.