Zeros of the Hankel function.

Question

I'm trying to find the zeros of the Hankel function (the first few will do) of the first kind $H^{(1)}_\nu(z) = J_\nu(z) + i Y_\nu(z)$ for complex argument $z$, does anyone know of a way to do this or some method or reference that would be useful

Answer 1

You might like to check out Complex zeros of cylinder functions

If you wan’t a quick and dirty approximation for large zeros then consider the analytic continuation $$\sin \left( v\pi \right)H_{v}^{1}\left( z{{e}^{m\pi i}} \right)=-\sin \left( \left( m-1 \right)v\pi \right)H_{v}^{1}\left( z \right)-{{e}^{-v\pi i}}\sin \left( mv\pi \right)H_{v}^{2}\left( z \right)$$ Then $$\sin \left( v\pi \right)H_{v}^{1}\left( z{{e}^{2\pi i}} \right)=-\sin \left( v\pi \right)H_{v}^{1}\left( z \right)-{{e}^{-v\pi i}}\sin \left( 2v\pi \right)H_{v}^{2}\left( z \right)$$ Hence the zeros of the first kind boils down to $$\frac{H_{v}^{1}\left( z \right)}{H_{v}^{2}\left( z \right)}=-{{e}^{-v\pi i}}\frac{\sin \left( 2v\pi \right)}{\sin \left( v\pi \right)}$$ Now taking the asymptotic expansion for large z we have $${{e}^{i\left( 2z-\tfrac{1}{2}\pi \right)}}\simeq -2\cos \left( v\pi \right)$$ Or $${{z}_{v,n}}\simeq \pi n-\frac{\pi }{4}-\frac{i}{2}\log \left( 2\cos \left( \pi v \right) \right)$$

we see at once that i have dropped a minus sign somewhere (the real part should be negative). however comparing the v=0 case of the above approximation to the table in the reference we have $${{z}_{0,40}}\simeq 124.8783079801-0.3465735902\text{i}$$ verses the 10-digit approximiation from the reference $${{z}_{0,40}}\simeq -124.8793089055-0.3465708127\text{i}$$ Ignoring my sloppy omission of a minus sign, as a first run it’s a very quick fix for two decimals.