I am trying to calculate $$\lim_{t\to\infty}e^{-zt}$$ where $z$ is complex and $Re(z)>0.$ This means that if we define $z=x+iy$, then $x>0.$
So
$$\lim_{t\to\infty}e^{-(x+iy)t}=\lim_{t\to\infty}e^{-xt}\lim_{t\to\infty}e^{-iyt}=e^{-\infty}\left(\cos(\pm\infty)+i\sin(\pm\infty)\right)=0\cdot\text{number}=0$$ Is this argument correct?
No, there is no such $\cos(\infty)$.
Rather, one can do like $|e^{-zt}|=|e^{-xt}|\cdot|e^{-iyt}|=e^{-xt}\rightarrow 0$ as $t\rightarrow\infty$ because $\text{Re}(z)=x>0$.