Is this proof of $\mathcal{L}\{\sin{at}\}=\frac{a}{s^2+a^2}$ valid? I found this proof of $\mathcal{L}\left\{\sin{at}\right\}=\frac{a}{s^2+a^2}$ on Proof Wiki:

This proof is way easier than others since it uses the linearity of the Laplace Transform. However, I am confused by the author's use of $\operatorname{Im}$. Isn't $\frac{a}{s^2+a^2}$ a complex number if $\Im{s}\neq0$? It seems like the author treats $s$ as a real number. Can we do the similar thing for complex $s$?

Inspired by the above, the conclusion can be proved in another way.
We know that
$$
{\cal L}\left\{ {{e^{iat}}} \right\} = \frac{1}{{s - ia}}
$$
and
$$
{\cal L}\left\{ {{e^{ - iat}}} \right\} = \frac{1}{{s + ia}}.
$$
With the help of Euler's Formula, $\sin at$ can be written as
$$
\sin at = \frac{1}{{2i}}\left( {{e^{iat}} - {e^{ - iat}}} \right),
$$
so
$$
{\cal L}\left\{ {\sin at} \right\} = \frac{1}{{2i}}{\cal L}\left\{ {{e^{iat}}} \right\} - \frac{1}{{2i}}{\cal L}\left\{ {{e^{ - iat}}} \right\} = \frac{1}{{2i}}\left( {\frac{{s + ia}}{{{s^2} + {a^2}}} - \frac{{s - ia}}{{{s^2} + {a^2}}}} \right) = \frac{a}{{{s^2} + {a^2}}}
$$
