Determine the image of the strip $S$ consisting of all points $z$ with $\frac{-\pi}{2}\lt Re(z) \lt \frac{\pi}{2}$ and $Im(z)>0$ under $w=i\sin z$

Question

$\color{green}{\text{transformation is}\space w=i\sin z}$

$$w=i\sin z = i\sin(x+iy)=\frac{1}{2}\left(e^{ix-y}-e^{-(ix-y)}\right)=-\cos(x)\sinh(y)+i\sin(x)\cosh(y)$$

$\therefore u = -\cos(x)\sinh(y) \tag{1}$

$\color{green}{\text{and }} v = \sin(x)\cosh(y) \tag{2}$

$\color{green}{\text{1.Transformation for }\space y>0 \space \text{i.e.} \space \operatorname{Im}(z)>0}$

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.

Answer 1

*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 ;) )

Answer 2

So firstly note that you've let $x$ and $y$ represent $\mathrm{Re}(z)$ and $\mathrm{Im}(z)$ respectively. Therefore since $\frac{-\pi}{2} < \mathrm{Re}(z) < \frac{\pi}{2}$, neither $\mathrm{sin}(x)$ nor $\mathrm{cos}(x)$ are constant over the region.

There is a simpler way to solve this. Go back to $w = u + \mathrm{i}v$, with $u = -\cos(x)\sinh(y)$ and $v = \sin(x)\cosh(y)$. This bit is correct.

Since $\frac{-\pi}{2} < \mathrm{Re}(z) = x < \frac{\pi}{2}$, so $ 0 < \cos(x) < 1$ and $-1 < \sin(x) < 1$. Also $\mathrm{Im}(z) = y > 0$, so $\cosh(y) > 1$ and $\sinh(y) > 0$.

Then we can look at how this bounds $u$ and $v$. We get $u = -\cos(x)\sinh(y) < 0$ as any negative quantity (as it is the negative of the product of two positive quantities, one unbounded) and $v = \sin(x)\cosh(y) \in \mathbb{R}$ as any real number (as it is a positive, negative, or zero multiple of any positive real number).

So $\mathrm{Re}(w) < 0$ and $\mathrm{Im}(w) \in \mathbb{R}$.