# from complex function to a series

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.

-
After tidying up your expression and using FullSimplify on it, I get zero. Is that what you expected? –  Verbeia May 20 '13 at 7:12
Its the Log of thing/thing, and Log[1]=0. So the series is pretty straightforward! –  bill s May 20 '13 at 7:17
sorry I edited the function right now –  user6921 May 20 '13 at 7:46
SeriesCoefficient[ Log[(1 - E^((I Pi (1 - a))/(b - a)) z)/(1 - E^(-((I Pi (1 - a))/(b - a))) z)], {z, 0, n}] // FullSimplify –  chris May 20 '13 at 8:54

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

-