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Adding exponentials

  • #1
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Homework Statement



How does [itex]\frac{e^{2iz}+2+e^{-2iz}}{4}=\frac{2}{4}[/itex]?

This is part of something more complex, but this should be enough information. I do not see how those exponentials cancel with each other to leave the [itex]\frac{2}{4}[/itex].

Thanks
 

Answers and Replies

  • #2
HallsofIvy
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All I can say is that [itex](e^{2iz}+ 2+ e^{-2iz})/4[/itex] is NOT equal to "2/4"! For example, if z= 0, the left side is (1+ 2+ 1)/4= 4/4= 1, not 2/4. And since normally you would write "1/2" rather than "2/4", I seems clear to me that you are missing something. What is true is that [itex]e^{2iz}+ 2+ e^{-2iz}= (e^{iz}+ e^{-iz})^2[/itex]. Does that help?
 
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  • #3
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Thanks. I knew I wasn't that bad at math to not know how to add exponentials.

It's from an example in my book. Here is the whole thing:

Prove that [itex]sin^{2}z+cos^{2}z=1[/itex]

[itex]sin^{2}z=(\frac{e^{iz}-e^{-iz}}{2i})^{2}=\frac{e^{2iz}-2+e^{-2iz}}{-4}[/itex]

[itex]cos^{2}z=(\frac{e^{iz}+e^{-iz}}{2})^{2}=\frac{e^{2iz}+2+e^{-2iz}}{4}[/itex]

[itex]sin^{2}z+cos^{2}z=\frac{2}{4}+\frac{2}{4}=1[/itex]
 
  • #4
3,812
92
Thanks. I knew I wasn't that bad at math to not know how to add exponentials.

It's from an example in my book. Here is the whole thing:

Prove that [itex]sin^{2}z+cos^{2}z=1[/itex]

[itex]sin^{2}z=(\frac{e^{iz}-e^{-iz}}{2i})^{2}=\frac{e^{2iz}-2+e^{-2iz}}{-4}[/itex]

[itex]cos^{2}z=(\frac{e^{iz}+e^{-iz}}{2})^{2}=\frac{e^{2iz}+2+e^{-2iz}}{4}[/itex]

[itex]sin^{2}z+cos^{2}z=\frac{2}{4}+\frac{2}{4}=1[/itex]
Are you still in a doubt? Check the expansion of ##\sin^2z## in the first step, there's a minus sign in the denominator.
 
  • #5
575
47
Are you still in a doubt? Check the expansion of ##\sin^2z## in the first step, there's a minus sign in the denominator.
I just tried multiplying the top and bottom of the [itex]sin^{2}z[/itex] expansion by -1 and then adding the two together, I get [itex]\frac{4}{4}[/itex], which is correct. The example makes it look like the exponentials are cancelled before the addition of the two. How did they do that?

Thanks.
 
  • #6
3,812
92
I just tried multiplying the top and bottom of the [itex]sin^{2}z[/itex] expansion by -1
What do you get when you do that?
 
  • #7
575
47
What do you get when you do that?
I got [itex]\frac{-e^{2iz}+2-e^{-2iz}}{4}[/itex] and then I added that with the cosine expansion and got 1. But the exponentials didn't disappear until I added the sine and cosine, at which point they cancelled. But in the book, they were gone before the addition of sine and cosine. How did they do that?

Thanks.
 
  • #8
3,812
92
But in the book, they were gone before the addition of sine and cosine.
They were never gone before the addition. Try to add them yourself.
 

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