Laplace transform of complex exponential

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The Laplace transform of a complex exponential function follows the same principle as that of a real exponential, expressed as L[f(t)] = 1/(s - a) for f(t) = e^{at}. When dealing with complex numbers, the transform remains L[f(t)] = 1/(s - complex_number), affirming that all numbers can be represented as complex. The discussion highlights that this method can also be applied to derive the Laplace transforms of sine and cosine functions. Specifically, the transforms yield L[cos(a x)] = s/(s^2 + a^2) and L[sin(a x)] = a/(s^2 + a^2). This confirms the validity of using complex numbers in Laplace transforms.
EvLer
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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.
 
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Definitely! You can demonstrate it by direct integration.
 
EvLer said:
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)
 
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