$\color{green}{\text{transformation is}\space w=i\sin z}$
$$w=i\sin z = i\sin(x+iy)=\frac{1}{2}\left(e^{ix-y}-e^{-(ix-y)}\right)=-\cos(x)\sinh(y)+i\sin(x)\cosh(y)$$
$\therefore u = -\cos(x)\sinh(y) \tag{1}$
$\color{green}{\text{and }} v = \sin(x)\cosh(y) \tag{2}$
$\color{green}{\text{1.Transformation for }\space y>0 \space \text{i.e.} \space \operatorname{Im}(z)>0}$
using $(1)$ and $(2)$ :
$\dfrac{v^2}{\sin^2(x)}-\dfrac{u^2}{\cos^2(x)} = \cosh^2(y)-\sinh^2(y) \tag{a}=1$
$\color{red}{\text{don't know if it is legal to do this but i now assume} \sin(x) \text{ and } \cos(x) \text{ are constant terms}}$
$\therefore \dfrac{v^2}{\alpha^2}-\dfrac{u^2}{\beta^2} = 1$ therefore line $y>1$ transforms into a hyperbola
$\color{green}{\text{2.Transformation for }- \pi /2 \lt \operatorname{Re}(z) < \pi/2 \space }$
using $(1)$ and $(2)$ same way as in $(a)$
$\dfrac{v^2}{\sinh^2(x)}+\dfrac{u^2}{\cosh^2(x)} = \cos^2(y)+\sin^2(y) \tag{b}=1$
$\therefore \dfrac{v^2}{A^2}+\dfrac{u^2}{B^2} = 1$
is this the correct way to do this transformation?, thank you for any help you can offer.
Best Answer
*small correction
" ... therefore line $y>\color{red}{0}$ transforms ..."
It's looking good, the last part would be an ellips. When $x$ is big enough it would start looking like a circle (since $\cosh x \approx \sinh x$ when $x$ big enough)
Also, sure it is legal to perform that first transformation. You just take $x$ constant to investigate the effect of a change in $y$. (a vertical line)
But then you would need to know what bounds come into play. As Michael also stated:
$$\begin{align} u &= -\cos x \sinh y\\ v & = \sin x \cosh y \end{align}$$
Note that $\sinh y \in {]0, +\infty[}$ and $\cosh y \in {]1, +\infty[}$ if $y>0$.
And $-\cos x \in {[-1,0[}$ and $\sin x \in {]-1,1[}$ if $x\in {\left]-\frac{\pi}{2}, \frac{\pi}{2}\right[}$
Looking for upper bounds, $u$ will always consist of a negative number multiplied with values which go from 0 to $\infty$. In conclusion: $u \in {]-\infty, 0[}$ while $v$ can be either positive and negative while taking all values. $v\in \mathbb{R}$.
End result
$w \in \{ z \in \mathbb{C}: \operatorname{Re}z < 0\} $
Note
I like your calculations too, they show a deeper understanding of the map. (nice!)
I'va added a drawing where you can see the map taking an effect on a horizontal like and a vertical line. You can download the GeoGebra document here: http://tinyurl.com/ndjsn4o so you could play around with the points itself.
(this might be a bit overkill as an answer though ;) )