# Bessel function with complex argument

So I understand that the bessel functions of the first kind are the ones that satisfy this equation:

$$x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+(x^2-\alpha^2)y = 0$$

and the result is a linear combination of the bessel functions of the first and second kind.

equation(1): $$A J_a(x) + B Y_a(x)$$

Now let: $x = iv$

$$\dfrac{dy}{dx} = \dfrac{dy}{dv} \dfrac{1}{i} \\ \dfrac{d^2y}{dx^2} = -\dfrac{d^2y}{dx^2}$$

Substituting in the original equation we get:

$$(iv)^2(-1)\frac{d^2y}{dv^2}+(iv)\dfrac{1}{i}\frac{dy}{dv}+((iv)^2-\alpha^2)y = 0 \\ v^2\frac{d^2y}{dv^2}+v\frac{dy}{dv}- (v^2+\alpha^2)y = 0$$

This is the equation which has solutions the modified bessel functions.

Is equation 1 with x = iv a solution to this equation? ( I think it is but not sure )

and the second equation is: Why is this then not true?

$$J_a(ix) = I_a(x) \\ Y_a(ix) = K_a(x)$$

## 1 Answer

Your computations are correct. Recall, however, that the space of solutions is two-dimensional: there are infinitely many ways to choose a basis there. We choose the basis as $J_\alpha$ and $Y_\alpha$, which are real-valued functions on the real axis.

Similarly, for the modified Bessel's equation we choose the basis $I_\alpha, K_\alpha$ of real-valued functions. These are linear combinations of $J_\alpha(ix)$ and $Y_\alpha(ix)$, which themselves are not real-valued.

Specifically, $$I_\alpha(x) = e^{-\alpha \pi i/2}J_\alpha(ix)$$ and $$K_\alpha(x) = \frac{\pi}{2}e^{(\alpha+1) \pi i/2}(J_\alpha(ix)+i Y_\alpha(ix))$$