Thread: Mapping involving hyperbolic functions

The question:
For the mapping f(z) = sinh(z), find and sketch the image of Re(z) = c

My attempt:
Unless I'm mistaken, Re(z) = C is just a line.

I'm having problems with the f(z) = sinh(z) part. I set f(z) = w, and used the definition of sinh(z) as follows:







Now I'm stuck. :/ Have I attempted this correctly, and if so, how do I proceed? Any help would be greatly appreciated!

Yes, it's correct.

So now that you know the real part is equal to a constant C

cos(y)(e^x - e^{-x}) = C

cos(y)(e^x - 1/e^x) = C

cos(y)(e^{2x} - 1) = Ce^x

cos(y) = Ce^x/(e^{2x} - 1)

y = arccos[Ce^x/(e^{2x} - 1)].

I'm sure you can sketch this.

Originally Posted by Glitch

That is right.





Now, for and , we have that varies in . So, the image of is .

For use that varies in .

Edited: I misread the question. I thought the problem only was to find the image of:

@Prove It

Are you sure? The solution is:

= 1; c != 0

Where did u and v come from?

Originally Posted by Prove It

Where did u and v come from?

Judging by past questions, the convention is:

w = u + iv
z = x + iy

We have







and we obtain an ellipse.





and we obtain the set