# Laplace transform of complex exponential

1. Aug 24, 2005

### 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.

2. Aug 25, 2005

### Tide

Definitely! You can demonstrate it by direct integration.

3. Aug 25, 2005

### lurflurf

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)

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