Trigonomentic mapping question

**The question**

For the mapping f(z) = sinh(z), find and sketch the image of Im(z) = d.

If I'm not mistaken, this is just a horizontal line in the z-plane through some constant d. With mapping questions involving Z, I usually try and write f(z) in terms of z, then substitute it into the equation.

However this one has me stumped. I tried:

Let f(z) = w

$\displaystyle w = \frac{e^z - e^{-z}}{2}$

$\displaystyle 2w = e^z - e^{-z}$

Now I'm unsure of how to write this in terms of z. Is this the correct approach? If so, how do I progress?

Thank you.

Re: Trigonomentic mapping question

$\displaystyle \displaystyle \sinh{(z)} = \sinh{(x + iy)} = \sinh{(x)}\cos{(y)} + i\cosh{(x)}\sin{(y)}$.

You need to plot the imaginary part of this equal to $\displaystyle \displaystyle d$, so

$\displaystyle \displaystyle \begin{align*} \cosh{(x)}\sin{(y)} &= d \\ \sin{(y)} &= \frac{d}{\cosh{(x)}} \\ y &= \arcsin{\left[\frac{d}{\cosh{(x)}}\right]} \end{align*}$

Now you need to plot this.