# Finding zeros of a complex Airy function

I want to find the zeros of the solution to this ODE

DSolve[{y''[x] - (I x - 2) y[x] == 0, y==0,y'==1}, y[x],x]


I use

ResourceFunction["Intercepts"]


and get a product of several hypergeometric functions. Are there alternative ways of getting the zeros?For instance, such as the Bessel zeros, or the sine cosine zeros at $$n\pi$$ and $$n\pi+\frac{\pi}{2}$$?

Thanks

• You could use something like the the technique in this answer to get zeroes within a range. May 28 at 16:48
• There is AiryAiZero just like BesselJZero, but Mma cannot solve Solve[BesselJ[1, x] == BesselJ[2, x], x, Method -> Reduce] or even Solve[BesselJ[1, x] == 1/10, x, Method -> Reduce], even though it can solve Solve[BesselJ[1, x] == 0 x, Method -> Reduce]. On a bounded interval, it can solve Solve[BesselJ[1, x] == BesselJ[2, x] && 0 < x < 10, x, Method -> Reduce]. May 28 at 17:16

Maybe something like this, with a bounded domain:

func = y[x] /.
First@
DSolve[{y''[x] - (I x - 2) y[x] == 0, y == 0, y' == 1},
y[x], x];

Solve[func == 0 && -1/10 < Re[x] < 4 && -1 < Im[x] < 2, x]
(*
{{x -> 0},
{x ->
Root[{AiryAi[-(-1)^(2/3) (-2 + I #1)] AiryBi[2 (-1)^(2/3)] -
AiryAi[2 (-1)^(2/3)] AiryBi[-(-1)^(2/3) (-2 + I #1)] &,
1.9742593 + 0.4276746 I}]},
{x ->
Root[{AiryAi[-(-1)^(2/3) (-2 + I #1)] AiryBi[2 (-1)^(2/3)] -
AiryAi[2 (-1)^(2/3)] AiryBi[-(-1)^(2/3) (-2 + I #1)] &,
3.4529450 + 1.0382391 I}]}}
*)


Root` objects are an exact representation of algebraic or, in this case, transcendental roots. They are numeric and may be calculated to arbitrary precision.