How does $e^{ikz}$ ensure that the solution goes zero at $\pm\infty$?

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Here is a Fourier transform question:

Find the solution $T(z,\theta,t)$ to the heat equation on a cylindrical surface:

$$\dot{T}=\alpha (\partial_z^2+R^2\partial_\theta^2)T$$

where $R$ and $\alpha$ are constants. Let the initial temperature profile be $T(z,\theta,0) = \frac{1}{1+z^2}$ and assume $T \rightarrow 0$ as $z \rightarrow \pm \infty$.

Separating variables $T=g(z)f(\theta)h(t)$ gives the following ODEs:

$$f''(\theta)=-n^2 f(\theta)$$

$$h'(t)=-\alpha\lambda h(t)$$

$$g''(z)=(-\lambda+(nR)^2) g(z)=-k^2g(z)$$

I am not sure how to make sure that when $z \rightarrow \pm \infty$, $g(z) \rightarrow 0$. The solution to the $g(z)$ equation in complex exponentials is:

$$g(z)=Ae^{ikz}+Be^{-ikz}$$

I read the solutions and it said to use $e^{ikz}$ only. But I am not sure how it satisfies the $z \rightarrow \pm \infty$ boundary condition. What is wrong with $e^{-ikz}$ here?

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If $T(z,\theta,t=0) = \frac{1}{1+z^2}$ then it looks like separation of variables may not be satisfied i.e., it could happen that $T$ is not of the form $g(z) f(\theta) h(t)$. If you want separation of variables then $T(z,\theta,t=0) = g(z) f(\theta) h(0)$. Hence your solution assuming separation of variables has to be of the form $T(z,\theta,t) = \frac{h(t)/h(0)}{1+z^2}$ with $h(0) \neq 0$.