Consider the Hankel function of the first kind $H^{(1)}_\nu (z)$. If I restrict $z$ to be a purely imaginary number of the form $ix$ where $x \in \mathbb{R}$ and $x>0$ and let $\nu =0$. We can see that $iH^{(1)}_0 (ix) > 0$. Using the standard recursion relationship for the derivatives of the Hankel (and Bessel) functions, we see that $\frac{\partial}{\partial x} \big(iH^{(1)}_0 (ix)\big) = H^{(1)}_1 (ix)$, and running some calculations, we see that $H^{(1)} _1 (ix) < 0$ for $x>0$.
If you continue taking the $x$ derivatives of $iH_0 ^{(1)} (ix)$, the sign of the derivatives alternates for $x>0$. My question is, how do I prove that for even $n$, $\frac{\rm d ^n}{\rm d x^n} \big(iH^{(1)}_0 (ix)\big) >0$ and for odd $n$, $\frac{\rm d ^n}{\rm d x^n} \big(iH^{(1)}_0 (ix)\big) <0$?
We can transform the Hankel function into a modified Bessel function using the correspondence (DLMF) for $x>0$ \begin{equation} i{H^{(1)}_{0}}\left(ix\right)=\frac{2}{\pi}K_{0}\left(x\right) \end{equation} Now, the $k$-th derivative can be obtained from the expression (DLMF) of that of the modified Bessel function: \begin{align} &K_{0}^{(k)}\left(x\right)=\\ &\frac{1}{2^{k}}\left(e^{-ik\pi}K_{-% k}\left(x\right)+{k\choose1}e^{-i(k-2)\pi}K_{-k+2}\left(x% \right)+{k\choose2}e^{-i(k-4)\pi}K_{-k+4}\left(x\right)+% \cdots+e^{ik\pi}K_{+k}\left(x\right)\right) \end{align} All phase factors of the sum equal $-1$ if $k$ is odd and $1$ if $k$ is even. As \begin{align} \frac{d^k}{dx^k}\left[ i{H^{(1)}_{0}}\left(ix\right)\right]&=\frac{2}{\pi}\frac{d^k}{dx^k}\left[K_{0}\left(x\right)\right] \end{align} Since $K_{-\nu}(x)=K_\nu(x)$ and $K_\nu(x)$ is positive, the sign of the $k$-th derivative is the same as $(-1)^k$ as expected.