2
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How can I express this complex function as a series?

Log[
 (1 - E^((I Pi (1 - a))/(b - a)) z)/
 (1 - E^(-((I Pi (1 - a))/(b - a))) z)
]

Where z is the complex number, a and b are real numbers, a < 1 and b > 1.

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4
  • $\begingroup$ After tidying up your expression and using FullSimplify on it, I get zero. Is that what you expected? $\endgroup$
    – Verbeia
    May 20, 2013 at 7:12
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    $\begingroup$ Its the Log of thing/thing, and Log[1]=0. So the series is pretty straightforward! $\endgroup$
    – bill s
    May 20, 2013 at 7:17
  • $\begingroup$ sorry I edited the function right now $\endgroup$
    – user6921
    May 20, 2013 at 7:46
  • 1
    $\begingroup$ SeriesCoefficient[ Log[(1 - E^((I Pi (1 - a))/(b - a)) z)/(1 - E^(-((I Pi (1 - a))/(b - a))) z)], {z, 0, n}] // FullSimplify $\endgroup$
    – chris
    May 20, 2013 at 8:54

1 Answer 1

1
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Let us define

   tt=Log[(1 - E^((I Pi (1 - a))/(b - a)) z)/(1 - E^(-((I Pi (1 - a))/(b - a))) z)];

Then, in order to get the Series expansion coefficient to any order

   coef= SeriesCoefficient[tt, {z, 0, n}] // FullSimplify

(* -((2 I sin((π (a-1) n)/(a-b)))/n) *)

Indeed

  tt2=Sum[coef[[1, 1, 1]] z^n, {n,1, Infinity}]-tt

is null.

If you want to check

    Series[tt2, {z, 0, 15}]

(* O[z^16] *) for instance.

Note that you could do the Taylor expansion by hand

 Table[1/n! D[tt, {z, n}] /. z -> 0 // 
 FullSimplify[#, Assumptions -> {a < 1, b > 1}] & // Together, {n, 
 1, 5}]//TableForm

and identify the prefactor of Sin[n(a-1)π/(a-b)] as -2I/n

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