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The ordinary differential equations arising from the separation of variables technique for the Laplace equation are:

\begin{align} &&\frac{1}{X}\frac{d^2 X}{d x^2}=k&&\text{and}&&\frac{1}{Y}\frac{d^2 Y}{d y^2}=-k&& \end{align}

where $X(x)Y(y)$ is the separable solutions. There is nothing that restricts the constant $k$ in principle to the positive real numbers, and so the solutions to these equations can, in theory, take on complex values for

$$X(x)=Ae^{\sqrt{k}x}+Be^{-\sqrt{k}x}$$

and

$$Y(y)=C\sin\sqrt{k}y+D\cos\sqrt{k}y$$

when $k$ is negative. I do not often encounter complex values in physics outside of quantum mechanics where they are a familiar part of the pedagogy, and am not sure how these complex solutions are interpreted physically? Are they discarded on the grounds that imaginary solutions don't correspond to actual potentials, or must either $\mathrm{Re}(V)$ or $\lvert\lvert V\rvert\rvert$ be considered?

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In classical electrodynamics, the potentials are real. There is no physical interpretation of nonreal potentials. When solving the Laplace equation, we have to restrict ourselves to real solutions.

That doesn't mean $k$ isn't allowed to be negative in the given expressions, but the final result $V(x,y) = X(x) Y(y)$ must be real. Note that if $m < 0$ then $\exp(\sqrt{m}x) = \cos(\sqrt{-m}x) + i \sin(\sqrt{-m}x)$. Similarly, $\sin(\sqrt{m}x)$ and $\cos(\sqrt{m}x)$ are linear combinations of $\exp(\sqrt{-m}x)$ and $\exp(-\sqrt{-m}x)$. So the solution you get is in the form

$$V(x, y) = A' e^{\sqrt{-m}x} \sin(\sqrt{-m}x) + B' e^{\sqrt{-m}x} \cos(\sqrt{-m}x) + C' e^{-\sqrt{-m}x} \sin(\sqrt{-m}x) + D' e^{-\sqrt{-m}x} \cos(\sqrt{-m}x)$$

But you would get those solutions anyway by considering linear combinations with $k = -m$ (which is positive), so you don't get anything new from considering negative $k$.

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