trig series can be summed with Euler's formula

[AlDanial]

Hi. I'm very impressed by sympy, terrific stuff. I recently tried to sum the series
sin( pi*(k-1)/N )
for k = 1..N but invoking .doit() just returns the original equation:

In [29]: from sympy import I, pi, symbols, Sum, cos, sin, exp
In [30]: N, k = symbols('N k')       
In [31]: Sum( sin(pi*(k-1)/N), (k, 1, N) ).doit()   
Out[31]: Sum(sin(pi*(k - 1)/N), (k, 1, N))

However, knowing exp(I*x) = cos(x) + I*sin(x), one can replace sin(pi*(k-1)/N) with exp(I*pi*(k-1)/N) and just keep the imaginary part of that sum. Thankfully sympy has no issue summing with exp():

In [35]: Sum( exp(I*pi*(k-1)/N), (k, 1, N) ).doit()   
Out[35]: Piecewise((N, Eq(exp(I*pi/N), 1)), ((exp(I*pi/N) - exp(I*pi/N)**(N + 1))/(-exp(I*pi/N) + 1), True))*exp(-I*pi/N)

so we just need to keep the imaginary parts of that answer. In my application I coded up the above equation numerically then just pulled out the ().imag part--worked great.

Is this a trick that can be coded into sympy's smarts so that the original series with the sine yields the answer like Imag(Piecewise((N, Eq(exp(I*pi/N), 1)), ((exp(I*pi/N) - exp(I*pi/N)**(N + 1))/(-exp(I*pi/N) + 1), True))*exp(-I*pi/N))? For the record I didn't arrive at this myself; my colleague Alain Sei showed me the technique and the sum which he derived himself.