Question on linear combinations of sines and cosine (complex analysis)

[arunma]

I have a question on complex analysis. Given a differential equation,

\dfrac{d^2 \psi}{dx^2} + k ^2 \psi = 0

we know that the general solution (before imposing any boundary conditions) is,

\psi (x) = A cos(kx)+B sin(kx).

Now here's something I don't quite understand. The solution,

\psi (x) = A'e^i ^k ^x + B'e^-^i ^k ^x

also works. I'm told that there's a way to rewrite one solution in terms of the other, and solve for the coefficients A and B in terms of A' and B'. But when I write the sines and cosines in terms of imaginary exponentials, I find that one set of coefficients must be complex valued. Is there any way to write \psi (x) in both ways, but keep all the coefficients real?

In case anyone's wondering, I'm asking because this expression is the solution to the differential equations that pop up on quantum mechanics problems on my PhD qualifier (which is in a month and a half).

 

[alyscia]

Yes, look up Euler's formula.
$e^{i\theta} = cos \theta + i sin \theta$.

 

[HallsofIvy]

alyscia, I don't believe that's his question. arunma's question is whether it is possible to write Acos(kx)+ Bsin(kx)= A'e^ikx+ B'e^-ikx with all four of A, B, A', and B' real. The answer to that is "no". If A and B are both real, then Acos(kx)+ Bsin(kx) is real for all x and so at least one of A' and B' must be non-real.

 

[arunma]

alyscia, I don't believe that's his question. arunma's question is whether it is possible to write Acos(kx)+ Bsin(kx)= A'e^ikx+ B'e^-ikx with all four of A, B, A', and B' real. The answer to that is "no". If A and B are both real, then Acos(kx)+ Bsin(kx) is real for all x and so at least one of A' and B' must be non-real.

Thanks for your help. I think this takes care of the confusion I've been having for the past couple weeks.