Why the imaginary part of $e^{-(a+jw)t}$ can't affect the limit as $t$ goes to infinity?

Question

I am learning the Fourier Transform. For a signal, $$ x(t) = e^{(-at)} * u(t), \; a>0 $$ we know the FT of it is $$ X(jw) = \frac{1}{a + jw} $$ During the computation of the FS, we substitute the $t$ with infinity and $0$ to get the result of the integration: $$ e^{-(a+jw)t} $$ How do we know when t goes to infinity the result goes to zero? I know if $$z = e^{-at}, \; a>0$$ $z = 0$ when $t$ goes to infinity. But the previous one has an imaginary part. Why the imaginary part can't affect the result when t goes to infinity? Thanks!

Answer 1

Welcome, Hao Wu! I assume that by $ j $ you mean the imaginary unit; it might be good to explain that in the question, since in math we usually use $ i $ for this.

Anyway, $ e ^ { - ( a + j w ) t } = e ^ { - a t } \big ( \cos ( w t ) - j \sin ( w t ) \big ) $, right? As $ t \to \infty $, both $ \cos ( w t ) $ and $ \sin ( w t ) $ remain between $ - 1 $ and $ 1 $. So we can write $ - e ^ { - a t } \leq e ^ { - a t } \cos ( w t ) \leq e ^ { - a t } $ for the real part and $ - e ^ { - a t } \leq - e ^ { a t } \sin ( w t ) \leq e ^ { - a t } $ for the imaginary part, and since both $ - e ^ { - a t } \to 0 $ and $ e ^ { - a t } \to 0 $ as $ t \to \infty $, it follows that both $ e ^ { - a t } \cos ( w t ) \to 0 $ and $ - e ^ { - a t } \sin ( w t ) \to 0 $ as well. Since both the real and imaginary parts of $ e ^ { - ( a + j w ) t } \to 0 $, this means that $ e ^ { - ( a + j w ) t } \to 0 $ itself.

Or more briefly: $$ e ^ { - ( a + j w ) t } = e ^ { - a t } \big ( \cos ( w t ) - j \sin ( w t ) \big ) = e ^ { - a t } \cos ( w t ) - e ^ { - a t } \sin ( w t ) \, j \underset { t \to \infty } \longrightarrow 0 [ - 1 , 1 ] - 0 [ - 1 , 1 ] j = [ 0 , 0 ] - [ 0 , 0 ] j = 0 - 0 j = 0 \text , $$ where $ [ a , b ] $ stands for any and all numbers between $ a $ and $ b $.