The analytic continuation of the first Hankel function is given by (abramowitz_and_stegun): $H^{(1)}_{n}(ze^{m\pi i})=(-1)^{mn-1}((m-1)H^{(1)}_{n}(z)+mH^{(2)}_{n}(z))$ where m,n are integers and z is a complex number.
My question is the following: For $m=\pm1$ I would expect to obtain the same result for $H^{(1)}_{n}(-z)$. But this does not seem to be the case:
For $m=+1$ we have: $$H^{(1)}_{n}(ze^{\pi i})=(-1)^{n-1}(H^{(2)}_{n}(z))$$ and for $m=-1$ we have $$H^{(1)}_{n}(ze^{-\pi i})=(-1)^{-n-1}(-2H^{(1)}_{n}(z)-H^{(2)}_{n}(z))$$ I would expect to have $H^{(1)}_{n}(ze^{\pi i})=H^{(1)}_{n}(ze^{-\pi i})$ but this leads to $H^{(1)}_{n}(z)+H^{(2)}_{n}(z)=0\rightarrow J_{n}(z)=0$ where $J_{n}(z)$ is a Bessel function defined as $J_{n}(z)=\frac{1}{2}(H^{(1)}_{n}(z)+H^{(2)}_{n}(z))$
I am sure there is a subtle difference between the two cases $m=\pm1$, but I can't find it. Any ideas? Thank you!