# Cosh z=0 where z= x + iy

1. Mar 13, 2006

### En_lizard

cosh z=0 where z= x + iy

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

thanx

2. Mar 14, 2006

### d_leet

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?

3. Mar 14, 2006

### dextercioby

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

That is addition theorems in hyperbolic trigonometry.

Daniel.

4. Mar 14, 2006

### Cexy

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.

5. Mar 14, 2006

### SGT

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

6. Mar 14, 2006

### matt grime

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.

7. Mar 19, 2006

### En_lizard

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

8. Mar 20, 2006

### matt grime

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

9. Mar 9, 2007

### 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?

10. Mar 9, 2007

### dextercioby

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