A simple question for numerical inverse Laplace Transform

Question

I recently met a problem when I tried to apply Laplace transform to a simple complex-valued function, for example \begin{equation} f(t) = \exp\left(-t(a+bi)\right) \end{equation} where $a,b,t>0$. The forward transform is simple: \begin{equation} F(s) = \frac{1}{s+(a+bi)} \end{equation}

However, when I tried to use numerical Laplace inversion with various Matlab algorithms (e.g. https://www.mathworks.com/matlabcentral/fileexchange/32824-numerical-inversion-of-laplace-transforms-in-matlab?s_tid=srchtitle ), none of those can inverse $F(s)$ to $f(t)$, especially when $a$ and $b$ are large (eg. 0.01). In this case f(t=2000)=0.3341 - 0.7475i, but the inversion gives 0.8187. Other inverters give various results. Actually, when I set $b=0$, which means that the imaginary part is dismissed, then the results are well matched. Can someone tell me why?

the following are Matlab codes for a test:

    clc 
    close all
    clear all
    
        
    a=10^-2*1i;
    b=10^-4;
    tmin=10;
    tmax=2000;
    tnum=2; % numbers of output values btw tmin and tmax. 
    
    X=exp(-tmax*(a+b)) % f(t)
    
    % ---------------------- Laplace inversion test-
   
    funcf=sprintf('1/(%e+%e+s)',a,b);
        
    % laplace inverison 
    [ti,ccf]=INVLAP(funcf,tmin,tmax,tnum) % Laplace inversion