Adding Exponentials: Solving \frac{e^{2iz}+2+e^{-2iz}}{4}=\frac{2}{4}

[leroyjenkens]

Homework Statement

How does $\frac{e^{2iz}+2+e^{-2iz}}{4}=\frac{2}{4}$?

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 $\frac{2}{4}$.

Thanks

 

[HallsofIvy]

All I can say is that $(e^{2iz}+ 2+ e^{-2iz})/4$ 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 $e^{2iz}+ 2+ e^{-2iz}= (e^{iz}+ e^{-iz})^2$. Does that help?

 

[leroyjenkens]

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 $sin^{2}z+cos^{2}z=1$

$sin^{2}z=(\frac{e^{iz}-e^{-iz}}{2i})^{2}=\frac{e^{2iz}-2+e^{-2iz}}{-4}$

$cos^{2}z=(\frac{e^{iz}+e^{-iz}}{2})^{2}=\frac{e^{2iz}+2+e^{-2iz}}{4}$

$sin^{2}z+cos^{2}z=\frac{2}{4}+\frac{2}{4}=1$

 

[Saitama]

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 $sin^{2}z+cos^{2}z=1$

$sin^{2}z=(\frac{e^{iz}-e^{-iz}}{2i})^{2}=\frac{e^{2iz}-2+e^{-2iz}}{-4}$

$cos^{2}z=(\frac{e^{iz}+e^{-iz}}{2})^{2}=\frac{e^{2iz}+2+e^{-2iz}}{4}$

$sin^{2}z+cos^{2}z=\frac{2}{4}+\frac{2}{4}=1$

Are you still in a doubt? Check the expansion of $\sin^2z$ in the first step, there's a minus sign in the denominator.

 

[leroyjenkens]

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 $sin^{2}z$ expansion by -1 and then adding the two together, I get $\frac{4}{4}$, which is correct. The example makes it look like the exponentials are canceled before the addition of the two. How did they do that?

Thanks.

 

[Saitama]

I just tried multiplying the top and bottom of the $sin^{2}z$ expansion by -1

What do you get when you do that?

 

[leroyjenkens]

What do you get when you do that?

I got $\frac{-e^{2iz}+2-e^{-2iz}}{4}$ 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.

 

[Saitama]

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.