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

[En_lizard]

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

 

[d_leet]

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?

 

[dextercioby]

How about using

\cosh\left(a+b\right) =...?

That is addition theorems in hyperbolic trigonometry.

Daniel.

 

[Cexy]

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.

 

[SGT]

How about using

\cosh\left(a+b\right) =...?

That is addition theorems in hyperbolic trigonometry.

Daniel.

Remember also that cosh(iy) = cos y and sinh(iy) = i sin y

 

[matt grime]

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.

 

[En_lizard]

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

that was a typo!
it never occurred to e that some people might think of as a stupidperson who'd say such a illogical thing.

 

[matt grime]

But which of the parts is the typo? that is what you're supposed to understand by my comment.

 

[En_lizard]

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) \pi/2
sinh x =0 : x=1
sin y =0 : y= 2k \pi/2

is it right?

 

[dextercioby]

Try to solve the "sinh x=0" again. You've got a wrong answer.