= Solution
For $m\geq0$, expand $e^{i\rho\cos\theta}=\exp[i\rho(e^{i\theta}+e^{-i\theta})/2]$. The lowest power producing the Fourier harmonic $e^{im\theta}$ is $(i\rho/2)^m/m!$. Meanwhile $J_m(\rho)\sim(\rho/2)^m/m!$. Matching these lowest powers gives $a_m=i^m$.
For negative order use $J_{-m}=(-1)^mJ_m$ and the equality of the positive and negative Fourier coefficients of this even function of $\theta$. This gives $a_{-m}=(-1)^mi^m=i^{-m}$. Therefore the <Jacobi-Anger expansion> is
$$
\boxed{e^{i\rho\cos\theta}=\sum_{m\in\mathbb Z}i^mJ_m(\rho)e^{im\theta}
=J_0(\rho)+2\sum_{m=1}^\infty i^mJ_m(\rho)\cos(m\theta).}
$$
The cosine form has multiplicity one for $m=0$ and two for every positive $m$. Those multiplicities are essential for matching the incident <plane wave>.
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