How can I find all w in C such that cos(z + w) = cos z for all z in C?

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Discussion Overview

The discussion revolves around finding all complex numbers \( w \) such that the equation \( \cos(z + w) = \cos z \) holds for all complex numbers \( z \). Participants explore the implications of this equation using trigonometric identities and seek to identify all possible values of \( w \).

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant initiates the discussion by asking how to find all \( w \) in \( \mathbb{C} \) that satisfy the equation for all \( z \).
  • Another participant suggests solving the equation in the reals as a preliminary step.
  • One participant proposes that \( w = 2\pi \) is an obvious solution and hints at other solutions without detailing them.
  • A later reply confirms that \( w = 2\pi n \) (where \( n \) is an integer) is a solution but questions whether there are additional solutions beyond this form.
  • Another participant inquires about the range of \( n \) in the context of \( 4\pi n \) and suggests that if \( n \) is an integer, then no new solutions are introduced.
  • One participant clarifies that while \( w = 4\pi n \) works for all \( z \), it does not encompass all possible \( w \) since \( 2\pi \) is also valid but not included in the set defined by \( 4\pi n \).

Areas of Agreement / Disagreement

Participants generally agree that \( w = 2\pi n \) and \( w = 4\pi n \) are solutions, but there is disagreement regarding whether these represent all possible solutions. The discussion remains unresolved regarding the completeness of the set of solutions.

Contextual Notes

The discussion highlights the need to clarify the definitions and ranges of \( n \) when considering the solutions, as well as the implications of the periodic nature of the cosine function.

Mark C
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Hi,

How would I do the following:

Find all w in C, such that cos(z + w) = cos z, for all z in C.

Of course I would need to use the identity cos z = (e^iz +e^-iz)/2, but I have trouble finding w for all z.

Thank you
 
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Just to get to "warmed"...Can u solve the equation into reals??

Daniel.
 
Err I mess about with the expression a bit using trig laws and din't get very far.

But erm wouldn't the obvious be w = 2*pi and by extention all the others you should know but I've edited it out just to give you something to do.
 
Yes, w=2(pi) is certainly one solution, and showing that w = 2(pi)n (n an integer), works for all z is easy. However, how could I know that there are no other w that do this? Meaning, I could also show w = 4(pi)n works for all z, but those are not all the w's possible.

Thanks
 
What's the range of the second "n",the one included in "4\pi n"??

If it's Z (like the one with "2 \pi n"),then you have to agree that in this case there's nor NEW solution... :wink:

Daniel.
 
If I show, that w = 4(pi)n works for all z (n an integer), then I do not find all such w that do this, for all z in C, because 2(pi), works also, but is not in the set 4(pi)n , n an integer. So, if I would claim that all w=4(pi)n works for all z, this is true, but, I do not find all such w.
 

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