Supposing that $\alpha>0$ and $x>0$, define the function: $$ f(x,\alpha) = x^2 \left[ K_{i\alpha+1}(x) K_{i\alpha-1}(x) - K_{i\alpha}(x)^2 \right] $$ where $K_{i \alpha}(x)$ is the modified Bessel function of the second kind of order $i \alpha$ and evaluated at the point $x$.
I am interested in a series expansion fo $f(x,\alpha)$ near $x=0$. I am thrown off because DLMF is unavailable due to the government shutdown (this is my usual go-to for facts about special functions). I am not able to find a series expansion for $K_{i \alpha}(x)$ near $x=0$ for an order $i \alpha$ being complex-valued. Is there a way of determining the above?
EDIT: Maybe I am misinterpreting this Wikipedia article. I quote a series expansion for $0 < |z| \ll \sqrt{1+b}$ where $$ K_{b}(z) \approx \frac{\Gamma(z)}{2} \left( \frac{2}{z} \right)^{b} $$ I thought it looks like this is only true for real $b$. Is this true for complex valued $b = i \alpha$? If this is true then I find I have $$ f(x,\alpha) \approx x^2 \frac{\Gamma(i\alpha + 1)\Gamma(i\alpha - 1)}{4} \left( \frac{2}{x} \right)^{i \alpha + 1}\left( \frac{2}{x} \right)^{i \alpha - 1} - x^2 \frac{\Gamma(i\alpha)^2}{4} \left( \frac{2}{x} \right)^{i\alpha } \left( \frac{2}{x} \right)^{i\alpha } $$
The correct expansion for $z\to 0$ is (see here) \begin{equation} K_\nu(z)=\frac{1}{2}\Gamma(\nu)\left( \frac{z}{2} \right)^{-\nu}\left( 1+O(z^2) \right)+\frac{1}{2}\Gamma(-\nu)\left( \frac{z}{2} \right)^{\nu}\left( 1+O(z^2) \right) \end{equation} for $\nu\notin\mathbb{Z}$. By keeping the leading term only, \begin{align} x^2 \left[ K_{i\alpha+1}(x) K_{i\alpha-1}(x) - K_{i\alpha}(x)^2 \right]&=\frac{x^2}{4}\left[\Gamma(1+i\alpha)\left( \frac{x}{2} \right)^{-i\alpha-1} \Gamma(1-i\alpha)\left( \frac{x}{2} \right)^{i\alpha-1}+O(1)\right]\\ &=\frac{\pi \alpha}{\sinh\pi\alpha}+O(x^2) \end{align}