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

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SUMMARY

The equation \(\frac{e^{2iz}+2+e^{-2iz}}{4}=\frac{2}{4}\) is incorrect as demonstrated by substituting \(z=0\), yielding a left side value of 1, not 0. The correct relationship is established through the identity \(e^{2iz}+2+e^{-2iz}=(e^{iz}+e^{-iz})^2\). This discussion clarifies the misunderstanding regarding the cancellation of exponentials in the context of the trigonometric identity \(\sin^2 z + \cos^2 z = 1\), where the correct expansions for \(\sin^2 z\) and \(\cos^2 z\) are provided.

PREREQUISITES
  • Understanding of complex exponentials, specifically \(e^{iz}\) and \(e^{-iz}\)
  • Familiarity with trigonometric identities, particularly \(\sin^2 z + \cos^2 z = 1\)
  • Knowledge of algebraic manipulation of fractions and expressions
  • Basic understanding of mathematical proofs and identities
NEXT STEPS
  • Study the derivation of Euler's formula \(e^{iz} = \cos z + i \sin z\)
  • Learn about the properties of complex numbers and their applications in trigonometry
  • Explore the concept of complex conjugates and their role in simplifying expressions
  • Investigate the proofs of other trigonometric identities using exponential forms
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Students studying complex analysis, mathematicians interested in trigonometric identities, and educators teaching advanced algebra or calculus concepts.

leroyjenkens
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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
 
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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?
 
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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 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
 
leroyjenkens said:
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.
 
Pranav-Arora said:
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.
 
leroyjenkens said:
I just tried multiplying the top and bottom of the sin^{2}z expansion by -1

What do you get when you do that?
 
Pranav-Arora said:
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.
 
leroyjenkens said:
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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