1. 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:

$w=sinh(z)$
$w = \frac{e^z - e^{-z}}{2}$
$w = \frac{e^{x+yi} - e^{-(x+yi)}}{2}$
$w = e^x(cos(y) + isin(y)) - e^{-x}(cos(y) - isin(y))$
$w = cos(y)(e^x - e^{-x}) + isin(y)(e^x+e^{-x})$

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

2. 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.

3. Originally Posted by Glitch
$w = cos(y)(e^x - e^{-x}) + isin(y)(e^x+e^{-x})$

That is right.

$\textrm{Re\;}w:\mathbb{C}\to \mathbb{R}$

$(\textrm{Re\;}w)(x+iy)=2\cos y\sinh x$

Now, for $y=0$ and $x\in\mathbb{R}$ , we have that $2\sinh x$ varies in $\mathbb{R}$. So, the image of $\textrm{Re\;}w$ is $\mathbb{R}$.

For $\textrm{Im}\;z$ use that $2\cosh x$ varies in $[2,+\infty)$ .

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

$\textrm{Re}\;w\;,\;\textrm{Im}\;w$

4. @Prove It

Are you sure? The solution is:

$(\frac{u}{sinh(c)})^2 + (\frac{v}{cosh(c)})^2$ = 1; c != 0

5. Where did u and v come from?

6. 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

7. We have

$u+iv=f(c+iy)\Leftrightarrow \begin{Bmatrix} u=2\cos y\sinh c\\v=2\sin y\cosh c\end{matrix}$

$(i)\quad c\neq 0$

$\dfrac{u^2}{(2\sinh c)^2}+\dfrac{v^2}{(2\cosh c)^2}=1$

and we obtain an ellipse.

$(ii)\quad c=0$

$u+iv=f(0+iy)\Leftrightarrow \begin{Bmatrix} u=0\\v=2\sin y\end{matrix}$

and we obtain the set

$\{0\}\times [-2,2]$