It is not that difficult to compute the branches of \( h \).
The branches of \( h(b) \) are simply the fixed points of \( b^z \). In this way you can compute the branches \( k=0,\dots,\infty \) via
\( h_k(b)=\lim_{n\to\infty}\log_{k,b}^{\circ n}(-1) \), where \( \log_{k,b}(z)=\frac{\log(z)+2\pi i k}{\log(b)} \).
This is however valid only for \( b\notin [e^{-e},e^{1/e}] \) as the numbering of the branches in this interval may be a bit different. For \( b=e^{1/e} \) the branch 0 and -1 (which are non-real) join and for slightly smaller \( b \) it splits again into two real branches.
The negative branches are simply the conjugates of the positive branches:\( h_{-k}(b)=\overline{h_{k-1}(b)} \).
Edit:
As it is so easy to compute \( h \) it is perhaps more appropriate to compute \( W \) from \( h \) by \( W(z)=z\cdot h(e^{-z}) \) instead of computing \( h \) from \( W \).
The branches of \( h(b) \) are simply the fixed points of \( b^z \). In this way you can compute the branches \( k=0,\dots,\infty \) via
\( h_k(b)=\lim_{n\to\infty}\log_{k,b}^{\circ n}(-1) \), where \( \log_{k,b}(z)=\frac{\log(z)+2\pi i k}{\log(b)} \).
This is however valid only for \( b\notin [e^{-e},e^{1/e}] \) as the numbering of the branches in this interval may be a bit different. For \( b=e^{1/e} \) the branch 0 and -1 (which are non-real) join and for slightly smaller \( b \) it splits again into two real branches.
The negative branches are simply the conjugates of the positive branches:\( h_{-k}(b)=\overline{h_{k-1}(b)} \).
Edit:
As it is so easy to compute \( h \) it is perhaps more appropriate to compute \( W \) from \( h \) by \( W(z)=z\cdot h(e^{-z}) \) instead of computing \( h \) from \( W \).