The Handbook of Mathematical Functions gives the following relation for modified Bessel functions of the first kind: $$I_\nu(\lambda z)=\lambda^{\pm\nu}\sum_{k=0}^\infty \frac{(\lambda^2-1)^k(\frac{1}{2}z)^k}{k!}I_{\nu\pm k}\left(z\right)$$ (A&S 9.6.51).
I am interested in the case $\lambda = -1,\ \nu = \frac{1}{6}$ (and my $z$ is a complex number). As $\left(\lambda^{2} - 1\right)^{k}$ reduces to $0$ for $k \ne 0$, I am actually dealing with $$I_{\nu}\left(- z\right)=(-1)^{\pm\nu}I_{\nu}\left(z\right).$$
My problem lies in picking the correct form for $(-1)^{\pm\nu}$ between $e^{-i\pi\nu}$ and $e^{i\pi\nu}$. As $\nu$ is not an integer, they are obviously not equivalent. I can numerically check which one fits, but I would like to have a more robust argument than that.
Cheers
I figured it out. From DLMF 10.25.2,
$$\frac{I_\nu(z)}{I_\nu(-z)} = \frac{z^{\nu}}{(-z)^{\nu}}.$$
As $ \mathrm{ph}(-z) = \begin{cases} \mathrm{ph}(z) + \pi & \text{if $\mathrm{ph}(z) \in (-\pi, 0]$} \\ \mathrm{ph}(z) - \pi & \text{if $\mathrm{ph}(z) \in (0, \pi]$} \end{cases} $,
$$ \frac{I_\nu(z)}{I_\nu(-z)} = \begin{cases} e^{-i\pi\nu} & \text{if $\mathrm{ph}(z) \in (-\pi, 0]$} \\ e^{+i\pi\nu} & \text{if $\mathrm{ph}(z) \in (0, \pi]$} \end{cases}. $$