I want to know about zeros of $$I_0(z)=\frac{1}{\pi}\int_0^{\pi}e^{z\cos(t)}dt=\sum_{m=0}^{\infty}\frac{(z^2/4)^{m}}{(m!)^2}.$$
From the formula $I_0(z)=J_0(iz)$ and the fact that $J_0$ has infinitely many real zeros, we know $I_0(z)$ has no real zeros(by series expansion) and infinitely many isolated imaginary zeros.
But is there any other zeros outside the imaginary line? If not, is it possible to have a lower bound on the modulus in the form(for $|Re(z)|\ge c$, say) $$|I_0(z)|\ge e^{-|z|}~?$$
Best Answer
The reality of zeroes of $J_k(z), k >-1$ (hence the fact that $I_0$ has only purely imaginary zeroes) is well known - for $J_0$ it is known since Fourier in the 1820's- because for $a \ne b$ one has the easily proven integral relation (true for any $J_k, k >-1$ actually, proof by differentiating both sides and using the differential equation of $J_k$):
$$\int_0^x t J_0(at)J_0(bt)dt=\frac{x}{a^2-b^2}(J_0(ax)J_0'(bx)-J_0(bx)J_0'(ax))$$
This vanishes at $x=1$ if $a,b$ are distinct zeroes of $J_0$ (and in general of $J_k, k >-1$); but assuming $a$ is non real zero, then $\bar a \ne a$ is also a zero, while $t J_0(at)J_0(\bar a t)=t| J_0(at)|^2, t \in [0,1]$ so the integral above for $x=1$ is positive being $\int_0^1 t | J_0(at)|^2dt >0$ hence the required contradiction,
For more details and possibly material about your second question check Watson classic treaty on Bessel Functions.