lambertw, all branches

[Albert Chan]

(01-22-2024 06:09 PM)Gil Wrote:  Then
For k=1, a=-1 —> im(a) = 0, ie im(a) is not < 0.
Consequently, small_imag=0.

But then you write
if small_imag: x = LN(-a) + (0 if A<.5 else T/2)
But a=-1 small_imag = 0
Then my question: for that case, what is the value for initial xo?

We are left with easy case, x = T = mpc(0,2*k*pi+arg(a))

p2> W(-1, 1, True)
(0.0 + 9.42477796076938j)
(-2.05355675220527 + 7.63609249528568j)
(-2.06229155975165 + 7.58864445610365j)
(-2.06227772959937 + 7.58863117846966j)
(-2.06227772959828 + 7.58863117847251j)
(-2.06227772959828 + 7.58863117847251j)