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Good evening, I'm working on this step coming from a differential equation. I have:

$A\cos(\frac{kL}{2})+Ai\sin(\frac{kL}{2})+B\cos(\frac{kL}{2})-Bi\sin(\frac{kL}{2})=0$

$(A+B)\cos(\frac{kL}{2})+(A-B)i\sin(\frac{kL}{2})=0$

This expression is set equal to:

$A\cos(\frac{kL}{2})+B\sin(\frac{kL}{2})=0$

$A$ and $B$ are constants (may be complex, not specified for this). $L$ is a real number. Obvioulsy this should be solved for variable $k$.

I can't explain which rules has been used to write down this step. Many thanks.

muserock92
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  • Are $A$ and $B$ real? How about $k$ - real? Integer? And $L$? – Deepak Sep 05 '19 at 10:13
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    @Deepak: Even if they are, we can't deduce that $A\cos(\frac{kL}{2})+B\sin(\frac{kL}{2})=0$. It looks like somebody made a mistake here. To take a wild guess, perhaps $A+B$ and $A-B$ should be $A+iB$ and $A-iB$? – TonyK Sep 05 '19 at 10:15
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    The only explanation I can think of is that they are using $A,B$ etc for generic complex constants. The symbols in the two equations are not supposed to be the same constants. – Kavi Rama Murthy Sep 05 '19 at 10:17
  • @TonyK True. I'm just clarifying the question. – Deepak Sep 05 '19 at 10:17
  • I've added some details to the question including the previous step. – muserock92 Sep 05 '19 at 10:20

1 Answers1

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We have: $$A\cos\left(\frac{kL}{2}\right)+Ai\sin\left(\frac{kL}{2}\right)=Ae^{i\frac{kL}{2}}$$ $$B\cos\left(\frac{kL}{2}\right)-Bi\sin\left(\frac{kL}{2}\right)=Be^{-i\frac{kL}{2}}$$ Looking at this I would assume that the step made was redefining one of the constants, since we cannot get from this line to the one stated

Henry Lee
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  • Many thanks for your reply. So I should have $Ae^{i\frac{kL}{2}}+Be^{-i\frac{kL}{2}}=0$. I still have troubles to understand if the result given by my book is correct or not. If it can help, this comes by the quantum physics "Particle in a Box" demonstration, but I posted here because I think is a math issue. – muserock92 Sep 05 '19 at 10:24