Laplace transform of complex exponential

[EvLer]

I just want to be sure I understand this correctly, usually L[f(t)] = 1/(s-a), where $$f(t) = e^{at}$$, but if it is a complex number would still be 1/(s - complex_number)?
techinically, i think it should be, since every number can be reprsented as complex number. Just want to be sure about this with Laplace transform.
thanks much.

 

[Tide]

Definitely! You can demonstrate it by direct integration.

 

[lurflurf]

I just want to be sure I understand this correctly, usually L[f(t)] = 1/(s-a), where $$f(t) = e^{at}$$, but if it is a complex number would still be 1/(s - complex_number)?
techinically, i think it should be, since every number can be reprsented as complex number. Just want to be sure about this with Laplace transform.
thanks much.

Yes and it is a nice way to find the laplace transforms of sin and cos.
L[cos(a x)+i sin(a x)]=L[exp(i x)]=1/(s-a i)=(s+a i)/(s^2+a^2)
hence (equating real and imaginary parts)
L[cos(a x)]=s/(s^2+a^2)
L[sin(a x)]=a/(s^2+a^2)