calculus and analysis - finding an argument of a complex number

What is the simplest way to find the argument of the following function? ((1 - E^((I π (1 - α))/(β - α)) z)/(1 - E^(-((I π (1 - α))/(β - α))) z)) as I tried the polar way, but it is so complicated and I didn't reach the result yet.

Thank you.

Answer

Here's one path... define your function and translate to trig form:

f[x_, z_] := ExpToTrig[((1 - 
   E^((I π (1 - α))/(β - α)) z)/(1 - 
   E^(-((I π (1 - α))/(β - α))) z))]//. 
        (π (1 - α))/(-α + β) -> x

using a replacement of a new real variable x for the more complicated expression involving alphas and betas. This gives f[x,z] as

 (1 - z Cos[x] - I z Sin[x])/(1 - z Cos[x] + I z Sin[x])

Now break this apart so that it has real denominator (using a function from the help files for the Conjugate function)

 toRealDenominator[rat_] := 
   With[{c = ComplexExpand[Conjugate[Denominator[rat]]]}, 
   Expand[Numerator[rat] c]/Expand[Denominator[rat] c]]

And so

Simplify[toRealDenominator[f[x, z]]]

gives a complicated expression. But, since we're only interested in the phase, and the denominator is real-valued, we can just use the Numerator. Hence

Simplify[Numerator[toRealDenominator[f[x, z]]]]

gives

(-1 + z Cos[x] + I z Sin[x])^2

Now this is something squared, so the phase is 2 times the phase of that something. Hence

ComplexExpand[-1 + z Cos[x] + I z Sin[x] //. z -> a + b I]

gives the answer

-1 + a Cos[x] - b Sin[x] + I (b Cos[x] + a Sin[x])

For any z (defined via a and b) and any x (defined via alpha and beta) this is a complex number in the form of A + B I and so you can find its phase using Arg or ArcTan.