Can sinh x = 0 and cosh z = 0 Be Solved Simultaneously?

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Discussion Overview

The discussion revolves around the simultaneous solutions of the equations sinh(x) = 0 and cosh(z) = 0, where z is expressed as a complex number z = x + iy. Participants explore the implications of these equations in the context of hyperbolic functions, particularly focusing on the conditions under which these equations can be satisfied.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that cosh(z) = 0 is impossible, questioning the validity of the equation under complex numbers.
  • Others argue that sinh(z) = 0 has trivial solutions, with one participant suggesting z = 1 as a solution, while another corrects this to z = 0.
  • There is a suggestion to utilize hyperbolic addition theorems to analyze the equations further.
  • One participant points out that the notion of cosh(z) being always positive is only valid when z is restricted to real numbers, which is not applicable in this context.
  • Confusion arises regarding the interpretation of terms like "solvable" and "impossible," leading to a clarification about the logical consistency of these terms.
  • A mathematical breakdown of cosh(x + iy) is presented, leading to conditions on y and x, but the correctness of the proposed solutions is questioned by another participant.

Areas of Agreement / Disagreement

Participants express differing views on the solvability of cosh(z) = 0, with some asserting it is impossible while others challenge this notion. There is no consensus on the correctness of the proposed solutions or the interpretations of the equations.

Contextual Notes

Some participants highlight the need to consider the real and imaginary parts of complex numbers when determining solutions, indicating that assumptions about the nature of z can affect the discussion. There are also unresolved mathematical steps in the proposed solutions.

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cosh z=0 where z= x + iy

we can solve this equation since its imposible coshz=0, is it right?

how about sinh=0?

thanx
 
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En_lizard said:
cosh z=0 where z= x + iy

we can solve this equation since its imposible coshz=0, is it right?

how about sinh=0?

thanx

sinh(z) = 0 is somewhat trivial z = 1 is an obvious solution.

And if z = x + iy how is cosh(z) = 0 impossible? Do you know the formula for the inverse hyperbolic cosine? Have you tried using it?
 
How about using

[tex]\cosh\left(a+b\right) =...?[/tex]

That is addition theorems in hyperbolic trigonometry.

Daniel.
 
d_leet said:
sinh(z) = 0 is somewhat trivial z = 1 is an obvious solution.
I'm fairly sure that you mean z=0 ;)

To the OP, you should use dexter's hint, and remember that a complex number is zero only when its real and imaginary parts are both zero.
 
dextercioby said:
How about using

[tex]\cosh\left(a+b\right) =...?[/tex]

That is addition theorems in hyperbolic trigonometry.

Daniel.
Remember also that cosh(iy) = cos y and sinh(iy) = i sin y
 
En_lizard said:
cosh z=0 where z= x + iy

we can solve this equation since its imposible coshz=0, is it right?

So something is 'sovable' but 'impossible'? that makes no sense.

In any case the notion that cosh(z) is always positive (greater than 1) is only true if we restrict to z a real number, which is expressly not what is happening here.
 
matt grime said:
So something is 'sovable' but 'impossible'? that makes no sense..
that was a typo!:frown:
it never occurred to e that some people might think of as a stupidperson who'd say such a illogical thing.:eek:
 
But which of the parts is the typo? that is what you're supposed to understand by my comment.
 
cosh (x+iy)=0
cosh x cosh iy + sinh x sinh iy = cosh x cos y + i sinh x siny =0

therefor:
cos y = 0 : y =(2n+1) [tex]\pi/2[/tex]
sinh x =0 : x=1
sin y =0 : y= 2k [tex]\pi/2[/tex]

is it right?
 
  • #10
Try to solve the "sinh x=0" again. You've got a wrong answer.
 

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