lambertw, all branches

[Albert Chan]

Gil's Trivia (previous posts) summary

x = (2n)*pi*I
a = x*e^x = x*cis(0) = x
k = (im(x) + arg(x) - arg(a)) / (2*pi) = (2n)*pi / (2*pi) = n

W_k((2k)*pi*I) = (2k)*pi*I

W₁(2*pi*I) = 2*pi*I
W_-1(-2*pi*I) = -2*pi*I      // conjugate symmetry

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x = (2n+1/2)*pi*I
a = x*e^x = x*cis(pi/2) = x*I

if n≥0 then k = ((2n+1/2)*pi + pi/2 - pi) / (2*pi) = n
if n<0 then k = ((2n+1/2)*pi − pi/2 − 0) / (2*pi) = n

W_k(-(2k+1/2)*pi) = (2k+1/2)*pi*I

W₁(-5/2*pi) = 5/2*pi*I
W_-1(3/2*pi) = -3/2*pi*I

We can use conjugate symmetry for version with RHS = -x

W_-k(-(2k+1/2)*pi - 0*I) = -(2k+1/2)*pi*I

W_-1(-5/2*pi - 0*I) = -5/2*pi*I
W₁(3/2*pi - 0*I) = 3/2*pi*I

But this required signed zero. I had another version without using signed zero (last formula)
Example, W₁(3/2*pi) = 3/2*pi*I too, because it matched last formula form.

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odd = 2n - sign(n)      // n ≠ 0

x = odd*pi*I = (2n-sign(n))*pi*I
a = x*e^x = x*cis(odd*pi) = −x

if n>0 then k = ((2n−1)*pi + pi/2 + pi/2) / (2*pi) = n
if n<0 then k = ((2n+1)*pi − pi/2 − pi/2) / (2*pi) = n

a = -x should have -0 real part, but here, it does not matter.

k = (odd+sign(odd))/2
W_k(-odd*pi*I) = odd*pi*I

W₁(-pi*I) = pi*I
W_-1(pi*I) = -pi*I      // conjugate symmetry

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For im(x) mod (2*pi) = 3/2*pi --> im(x)/pi = odd + 1/2

x = (odd+1/2)*pi*I = (2n-sign(n)+1/2)*pi*I      // n ≠ 0
a = x*e^x = x*cis(3/2*pi) = x/I                        // im(a) = +0

if n>0 then k = ((2n-1/2)*pi + pi/2 - 0) / (2*pi) = n
if n<0 then k = ((2n+3/2)*pi - pi/2 - pi) / (2*pi) = n

k = (odd+sign(odd))/2
W_k((odd+1/2)*pi) = (odd+1/2)*pi*I

W_-2(-5/2*pi) = -5/2*pi*I
W₁(3/2*pi) = 3/2*pi*I