Complex Analysis - Solving Complex Trig functions

[NewtonianAlch]

Homework Statement

Now, I know there's two ways to go about this and it seems everywhere I look around on the web people are solving it in a way I think that seems longer, harder and more prone to mistakes in exams. It involves using the exponential identities and taking logs. I was shown another way, but unfortunately I haven't quite got the grasp of it.

sinh z = 0

sinh (x + iy) = sinh(x)cos(y) + i cosh(x)sin(y) = 0

So

1) sinh(x)cos(y) = 0
2) cosh(x)sin(y) = 0

(1) Either x = 0 or y = ± Pi/2
(2) cosh (x) is never 0, so therefore x is not 0. Hence y = 0

This is where I'm stuck, I do not know how to go from here.

The answer is z = i*k*Pi

 

[nebffa]

You have two simultaneous equations. You've correctly deduced that cosh(x) is never 0, so therefore in equation 2 divide both sides by cosh(x) to get

sin(y) = 0

which occurs when y = k*Pi, where k is an element of the integer set

Ok so we now know y! Let's look at x. Now you use your first equation:

at y = k*Pi, cos(y) = 1 or -1

so divide both sides by cos(y) and get

sinh(x) = 0, therefore x = 0

so what we have is z = x + iy = i * k * Pi.

This method is a lot more simple, it just involves working through the simultaneous equations carefully. Where you started to go wrong was when you assumed y just has to equal 0, rather than the general solution of k * Pi

 

[NewtonianAlch]

Hmm, interesting. I think I'm starting to see it now.

So suppose I have cosh z = 1

cosh (x + iy) = cosh(x)cos(y) + i sinh(x)sin(y) = 1

cosh(x)cos(y) = 1
sinh(x)sin(y) = 0

cosh(x) = 1 only when x = 0, and cos (y) = 1 for 2*k*Pi

So now we already have values for x and y, substituting this into the imaginary component, validates that.

So z = i*2*k*Pi

Is my reasoning correct? Thank you for your help by the way.

 

[NewtonianAlch]

Stuck again now!

cosh z = 2i

I posted this question earlier but back then I only knew the exponential and log method.

What I've got so far:

cos (x + iy) = cosh(x)cos(y) + i sinh(x)sin(y) = 2i

1) cosh(x)cos(y) = 0
2) sinh(x)sin(y) = 2i

From equation 1, cosh x cannot be zero. For cos (y) = 0, y is periodic so y = (Pi/2 + kPi)

Now this is where I'm stuck on how to deal with that in equation 2.

The final answer given is (-1)^k arcsinh(2) + i(kPi +Pi/2)

I can see how that answer for y I got in equation 1 has something to do with the final answer.

Edit:

I'm thinking that if y = (Pi/2 + kPi), this would mean sin(y) = (-1)^k

So dividing sin (y) in equation 2, we get sinh(x) = (-1)k 2i

Taking arcsinh of this now means arcsinh((-1)^k 2i)

Does this mean I can take out the (-1)^k ? I still haven't figured out what to do with the i hanging around.

 

[clamtrox]

cosh(x) = 1 only when x = 0, and cos (y) = 1 for 2*k*Pi

That's true but cosh(x) cos(y) = 1 holds for other values as well, for example cosh(x)=2, cos(y) = 1/2.

2) sinh(x)sin(y) = 2i

that should be 2, not 2i

 

[NewtonianAlch]

That's true but cosh(x) cos(y) = 1 holds for other values as well, for example cosh(x)=2, cos(y) = 1/2.



that should be 2, not 2i

That would explain it then, thanks!