Let $z=\alpha+i\beta$ and $s=x+iy$ be complex numbers.
Consider the calculation of the Laplace transform of $e^{-zt}$.
$$\mathcal{L}(e^{-zt})=\int_{0^-}^\infty e^{-(x+\alpha+i(\beta+y))t}dt$$
Suppose $x+\alpha=0$. That is $x=-\alpha$ which means $\text{Re}(s)=-\text{Re}(z)$.
Then the integral becomes
$$\mathcal{L}(e^{-zt})=\int_{0^-}^\infty e^{-i(\beta+y)t}dt$$
$$=\left . \frac{e^{-i(\beta+y)t}}{i(\beta+y)}\right |_0^\infty$$
$$=\left .\frac{1}{\beta+y}e^{-i\left (\frac{\pi}{2}+(\beta+y)t\right)}\right |_0^\infty$$
This integral does not go to $\pm \infty$.
What is the technical reason why this integral diverges?